Interpret rate of change and initial value

Learning goals

Identifies rate of change from a graph, table, or equation.
Identifies the initial value or y-intercept.
Interprets both quantities with units in context.
Uses them to reconstruct or compare linear models.

Difficulty: 1 – 5

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Interpret rate of change and initial value 1

For each linear function, identify the slope $m$ and y-intercept $b$. Then give the y-intercept point $(0, b)$. a) $f(x) = -4x + 7$ b) $g(x) = \frac{2}{5}x$ c) $h(x) = 12 - 3x$ d) $k(x) = -5$

Hints

- Compare each equation with $y = mx + b$. - Pay attention to the sign of each coefficient. - What is the slope of a constant function? - If there is no constant term, what is the y-intercept? - You may need to reorder terms before identifying $m$ and $b$.

Solution

1. For $f(x) = -4x + 7$, $m = -4$ and $b = 7$, so the y-intercept is $(0, 7)$. 2. For $g(x) = \frac{2}{5}x$, $m = \frac{2}{5}$ and $b = 0$, so the y-intercept is $(0, 0)$. 3. Rewrite $h(x) = 12 - 3x$ as $h(x) = -3x + 12$. Then $m = -3$ and $b = 12$, so the y-intercept is $(0, 12)$. 4. For $k(x) = -5$, the slope is $0$ and $b = -5$, so the y-intercept is $(0, -5)$.

Answer

a) $m = -4$, $b = 7$, $(0, 7)$ b) $m = \frac{2}{5}$, $b = 0$, $(0, 0)$ c) $m = -3$, $b = 12$, $(0, 12)$ d) $m = 0$, $b = -5$, $(0, -5)$

Interpret rate of change and initial value 2

The graph shows how many liters of water are in a container after $t$ minutes of filling. How much water was in the container when filling began? Explain what you read from the graph.

Hints

- The starting value happens before any minutes have passed. - Find the height of the line where the horizontal input is zero.

Solution

1. Filling begins at $t=0$. 2. The line meets the vertical axis at $8$, so the container starts with $8$ liters.

Answer

$8$ liters, the value at time $0$.

Interpret rate of change and initial value 3

Devin's savings, in dollars, are modeled by $S(w)=18+4w$, where $w$ is the number of weeks. What does the number $4$ mean in this situation? Include units.

Hints

- Notice which number is multiplied by the number of weeks. - Decide whether it represents a starting amount or a weekly change.

Solution

1. The coefficient of $w$ gives the change in savings for one more week. 2. Savings increase by $\$4$ each week.

Answer

Devin saves an additional $\$4$ per week.

Interpret rate of change and initial value 4

Camila is knitting a scarf. Its length is measured during an afternoon. <table><tr><th>Time (hours)</th><td>$0$</td><td>$2$</td><td>$4$</td></tr><tr><th>Length (in.)</th><td>$9$</td><td>$13$</td><td>$17$</td></tr></table> How long was the scarf at the start? Which entry tells you?

Hints

- Consider which column represents the time before the afternoon's knitting begins. - Match that time with the scarf's length in the same column.

Solution

1. The starting time is $0$ hours. 2. The table shows a length of $9$ inches at time $0$, so the scarf started at that length.

Answer

$9$ inches, from the column at $0$ hours.

Interpret rate of change and initial value 5

An elevator's height above the ground is modeled by $H(t)=60-3t$, where height is in feet and time is in seconds. What does the negative rate $-3$ mean for the elevator?

Hints

- Notice whether the number multiplying time is positive or negative. - A negative change in height means movement in which direction?

Solution

1. The coefficient of time is $-3$ feet per second. 2. The negative sign means the height decreases by $3$ feet each second.

Answer

The elevator descends by $3$ feet each second.

Interpret rate of change and initial value 6

A line $g$ is defined by $g(x) = 1.5x - 3$. a) Find the slope $m$ and the y-intercept of $g$. b) A second line $h$ is parallel to $g$ and passes through the origin. Write the equation of $h$. c) A third line $k$ has the same slope as $g$, but its y-intercept is $5$ units higher. Write the equation of $k$.

Hints

- What must be true about the slopes of parallel lines? - What are the coordinates of the origin? - Moving a y-intercept upward changes which parameter in $y = mx + b$?

Solution

1. From $g(x) = 1.5x - 3$, the slope is $m = 1.5$ and the y-intercept is $(0, -3)$. 2. A parallel line has the same slope. Since $h$ passes through the origin, its y-intercept is $0$. Thus $h(x) = 1.5x$. 3. Raising the y-intercept by $5$ changes it from $-3$ to $2$. Thus $k(x) = 1.5x + 2$.

Answer

a) $m = 1.5$; y-intercept $(0, -3)$ b) $h(x) = 1.5x$ c) $k(x) = 1.5x + 2$

Interpret rate of change and initial value 7

The linear function is $f(x)=2x-3$. Describe the new slope $m$ and y-intercept $b$, and write the new equation in each case. a) The graph shifts up $5$ units. b) The line becomes horizontal but keeps the same y-intercept. c) The line passes through the origin but keeps the same slope.

Hints

- Locate the slope and starting value in the original rule. - For each change, check whether the line's tilt or its crossing of the y-axis moves. - Test your new equation against the condition in that part.

Solution

1. Shifting up changes the y-intercept from $-3$ to $2$ but leaves slope $2$: $y=2x+2$. 2. A horizontal line has slope $0$ and keeps y-intercept $-3$: $y=-3$. 3. A line through the origin has intercept $0$ and keeps slope $2$: $y=2x$.

Answer

a) $m=2$, $b=2$; $y=2x+2$. b) $m=0$, $b=-3$; $y=-3$. c) $m=2$, $b=0$; $y=2x$.

Interpret rate of change and initial value 8

Consider the five linear functions: $f_1(x) = \frac{2}{5}x + 3$ $f_2(x) = 0.4x - 1$ $f_3(x) = 2.5x + 3$ $f_4(x) = \frac{5}{2}x - 4$ $f_5(x) = -0.4x + 3$. a) Which pairs of lines are parallel? Explain your reasoning. b) Which lines have the same y-intercept? Give the common point.

Hints

- What do equal slopes tell you about two lines? - Which term in $y = mx + b$ gives the y-intercept? - Converting equivalent fractions and decimals can make the comparisons easier.

Solution

1. The slopes are $\frac{2}{5} = 0.4$, $0.4$, $2.5$, $\frac{5}{2} = 2.5$, and $-0.4$. 2. Lines with equal slopes are parallel, so $f_1$ is parallel to $f_2$, and $f_3$ is parallel to $f_4$. 3. Functions $f_1$, $f_3$, and $f_5$ all have y-intercept $3$, so their graphs pass through $(0, 3)$.

Answer

a) $f_1 \parallel f_2$ and $f_3 \parallel f_4$ b) $f_1$, $f_3$, and $f_5$; common y-intercept $(0, 3)$

Interpret rate of change and initial value 9

The linear function $f$ is given by $f(x) = 1.2x - 3$. You want to use a slope triangle to draw its graph. Both axes use the same scale, so one grid square or one inch represents the same number of units on each axis. Starting at $P(0, -3)$, determine which moves correctly locate another point on the graph: 1. 5 units right and 6 units up 2. 10 grid squares right and 12 grid squares up 3. $2$ inches right and $2.4$ inches up 4. 6 units right and 5 units up 5. 5 units left and 6 units down

Hints

- What does the coefficient of $x$ tell you about the slope? - How can you compare a horizontal change and a vertical change to the slope? - Does the measurement unit matter when both axes use the same scale? - What happens to the signs of $\Delta x$ and $\Delta y$ when you move left and down?

Solution

1. The slope is $m = 1.2$, so a correct move must satisfy $\frac{\Delta y}{\Delta x} = 1.2$. 2. For move 1, $\frac{6}{5} = 1.2$, so it is correct. 3. For move 2, $\frac{12}{10} = 1.2$, so it is correct. 4. For move 3, $\frac{2.4}{2} = 1.2$, so it is correct. 5. For move 4, $\frac{5}{6} \approx 0.83$, so it is not correct. 6. For move 5, $\frac{-6}{-5} = 1.2$, so it is correct.

Answer

1. Correct 2. Correct 3. Correct 4. Incorrect 5. Correct

Interpret rate of change and initial value 10

Leen wants to graph $f(x) = -\frac{3}{4}x + 2$. Starting at the y-intercept, Leen considers three ways to make a slope triangle: 1. “Move $4$ grid squares right and $3$ grid squares down.” 2. “Move $1$ grid square right and $0.75$ grid square down.” 3. “Move $8$ grid squares left and $6$ grid squares up.” Evaluate all three methods. Which ones lead to the correct graph? Justify your answer with calculations.

Hints

- Identify the slope from the equation. - For each proposed move, compare the signed vertical change with the signed horizontal change. - Moving left makes the horizontal change negative, while moving down makes the vertical change negative.

Solution

1. The slope is $m = -\frac{3}{4} = -0.75$. 2. Method 1 gives $\frac{-3}{4} = -0.75$, so it is correct. 3. Method 2 gives $\frac{-0.75}{1} = -0.75$, so it is mathematically correct, although a fractional grid square may be harder to plot exactly. 4. Method 3 gives $\frac{6}{-8} = -0.75$, so it is correct.

Answer

All three methods are mathematically correct. Methods 1 and 3 are easiest to plot accurately on standard graph paper because they use whole grid squares.

Interpret rate of change and initial value 11

Consider the linear equation $\frac{1}{2}y-x=1$. a) Rewrite it in the form $y=mx+b$. b) State the rate of change and the initial value of the resulting function. c) If $x$ increases by $3$, by how much does $y$ change? Explain using the rate of change. d) Determine algebraically whether $P(5,12)$ lies on the graph.

Hints

- Isolate $y$ first so the coefficient and constant term have their standard linear-function meanings. - Connect a change in input to a change in output through the rate $m$. - Test the point only after you have identified the function rule.

Solution

1. Add $x$ to both sides: $\frac{1}{2}y=x+1$. Multiply by $2$: $y=2x+2$. 2. The rate of change is $m=2$, and the initial value is $b=2$. 3. A change of $3$ in $x$ changes $y$ by $m\cdot3=2\cdot3=6$. 4. At $x=5$, the function gives $y=2\cdot5+2=12$, so $P(5,12)$ lies on the graph.

Answer

a) $y=2x+2$ b) Rate of change: $2$; initial value: $2$ c) $y$ increases by $6$. d) Yes.

Interpret rate of change and initial value 12

Find the x-intercept and y-intercept of the linear function $y = \frac{2}{5}x + 4$.

Hints

- What is the x-coordinate of every point on the y-axis? - What is the y-coordinate of every point on the x-axis? - Substitute the appropriate zero and solve.

Solution

1. For the y-intercept, set $x = 0$. Then $y = 4$, so the y-intercept is $(0, 4)$. 2. For the x-intercept, set $y = 0$: $0 = \frac{2}{5}x + 4$. Then $-4 = \frac{2}{5}x$, so $x = -10$. The x-intercept is $(-10, 0)$.

Answer

x-intercept: $(-10, 0)$; y-intercept: $(0, 4)$

Interpret rate of change and initial value 13

The graph shows a youth group's hike. Time since departure is on the x-axis, and distance from the lodge is on the y-axis. a) How far is rest stop A from the lodge, and how long does the group rest there? b) Find the group's speed on the first segment, from $0$ to $2$ hours. c) When does the group begin returning, and how long does the entire outing last? d) Find the average speed on the return trip.

Hints

- A horizontal segment means the group's distance from the lodge is unchanged. - Speed is distance divided by time. - The return begins when the graph starts decreasing. - Use the distance at the start of the return segment to determine the return speed.

Solution

1. Rest stop A is $4$ miles from the lodge. The horizontal segment from $t=2$ to $t=3$ shows a $1$-hour rest. 2. On the first segment, the speed is $\frac{4}{2}=2\,\text{mph}$. 3. The return trip begins at $t=5$ hours. The graph returns to distance $0$ at $t=6$, so the outing lasts $6$ hours. 4. The group returns $6$ miles in $1$ hour, so its average return speed is $6\,\text{mph}$.

Answer

a) $4$ miles; $1$ hour b) $2\,\text{mph}$ c) The return begins after $5$ hours; the outing lasts $6$ hours. d) $6\,\text{mph}$

Interpret rate of change and initial value 14

The graphs of four linear functions, a, b, c, and d, are shown. Match each graph with the correct equation. 1) $y = -2x - 1$ 2) $y = \frac{1}{2}x - 2$ 3) $y = 2x + 1$ 4) $y = -\frac{1}{2}x + 3$ 5) $y = 2x - 1$ 6) $y = -\frac{1}{2}x - 2$ There are more equations than graphs, so two equations will not be used.

Hints

- First identify where each graph crosses the y-axis. - Use a slope triangle to find the rise and run for each graph. - Check whether each line rises or falls from left to right. - Compare the slope and y-intercept you found with the given equations.

Solution

1. Graph a has y-intercept $1$. It rises $2$ units for each $1$ unit to the right, so its slope is $2$. Therefore, a matches $y = 2x + 1$, equation 3. 2. Graph b has y-intercept $3$. It falls $1$ unit for every $2$ units to the right, so its slope is $-\frac{1}{2}$. Therefore, b matches $y = -\frac{1}{2}x + 3$, equation 4. 3. Graph c has y-intercept $-1$. It falls $2$ units for each $1$ unit to the right, so its slope is $-2$. Therefore, c matches $y = -2x - 1$, equation 1. 4. Graph d has y-intercept $-2$. It rises $1$ unit for every $2$ units to the right, so its slope is $\frac{1}{2}$. Therefore, d matches $y = \frac{1}{2}x - 2$, equation 2.

Answer

a: 3) $y = 2x + 1$; b: 4) $y = -\frac{1}{2}x + 3$; c: 1) $y = -2x - 1$; d: 2) $y = \frac{1}{2}x - 2$

Interpret rate of change and initial value 15

The graph shows how the height $h$ of a burning candle, in inches, decreases over time $t$, in hours. a) How tall was the candle at the start? b) Find the candle's burn rate in inches per hour. c) After how many hours is the candle completely burned down?

Hints

- The initial height is the graph's y-intercept. - Use two points to find the rate of change. - The candle is burned down when its height is $0$.

Solution

1. At $t=0$, the graph has a height of $9$ inches, so the candle's initial height is $9\,\text{in.}$. 2. Using $(0, 9)$ and $(4, 3)$, the slope is $m=\frac{3-9}{4-0}=-1.5$. The height decreases by $1.5\,\text{in./h}$, so the burn rate is $1.5\,\text{in./h}$. 3. The candle is completely burned down when $h=0$. Solve $0=-1.5t+9$ to get $t=6$.

Answer

a) $9\,\text{in.}$ b) $1.5\,\text{in./h}$ c) $6$ hours

Interpret rate of change and initial value 16

A battery's charge level is recorded once an hour. <table><tr><th>Time (h)</th><td>$0$</td><td>$1$</td><td>$2$</td><td>$3$</td><td>$4$</td><td>$5$</td><td>$6$</td><td>$7$</td><td>$8$</td></tr><tr><th>Charge level</th><td>$80\%$</td><td>$60\%$</td><td>$40\%$</td><td>$20\%$</td><td>$10\%$</td><td>$40\%$</td><td>$70\%$</td><td>$90\%$</td><td>$100\%$</td></tr></table> a) What is the starting charge level, and what does “starting” mean on this table? b) Find the average rate of change from hour $0$ to hour $3$, in percentage points per hour. Explain the sign. c) Find the average rate from hour $4$ to hour $7$. Round to the nearest tenth and explain the sign.

Hints

- Find the row entry at time zero before comparing later values. - For each time interval, think about the charge change and the hours that pass. - What does the sign of a change tell you about the battery level?

Solution

1. At $t=0$, the recorded charge is $80\%$. This is the starting charge level. 2. From hour $0$ to hour $3$, the average rate is $\frac{20-80}{3-0}=-20$ percentage points per hour. The negative sign means the charge drops. 3. From hour $4$ to hour $7$, the average rate is $\frac{90-10}{7-4}=\frac{80}{3}\approx26.7$ percentage points per hour. The positive sign means the charge rises on average.

Answer

a) $80\%$, the level at hour $0$ b) $-20$ percentage points per hour; the charge falls. c) About $+26.7$ percentage points per hour; the charge rises on average.

Interpret rate of change and initial value 17

Three linear functions are given. Match each function with graph p, q, or r, and briefly justify each match using the slope and y-intercept. - $a(x) = -\frac{2}{3}x + 1$ - $b(x) = \frac{3}{2}x - 2$ - $c(x) = -2x$

Hints

- The y-intercept is the value of $y$ when $x = 0$. Where does each graph cross the y-axis? - A negative slope means a line falls from left to right. Compare the steepness of the two decreasing lines. - A linear function with no constant term passes through $(0, 0)$.

Solution

1. For $a(x) = -\frac{2}{3}x + 1$, the y-intercept is $1$ and the slope is negative with magnitude less than $1$. Graph p has those features, so p represents $a$. 2. For $b(x) = \frac{3}{2}x - 2$, the y-intercept is $-2$ and the slope is positive. Graph q has those features, so q represents $b$. 3. For $c(x) = -2x$, the y-intercept is $0$, so the line passes through the origin, and its negative slope is steeper than the slope of $a$. Graph r has those features, so r represents $c$.

Answer

p: $a(x) = -\frac{2}{3}x + 1$; q: $b(x) = \frac{3}{2}x - 2$; r: $c(x) = -2x$

Interpret rate of change and initial value 18

Match each equation with the correct graph. Briefly justify each match using the slope and y-intercept. 1) $y = -2x + 3$ 2) $y = \frac{1}{3}x - 2$ 3) $y = x + 1$

Hints

- Identify the y-intercept in each equation and compare it with the graph. - Check whether each line rises or falls from left to right. - Compare the steepness of each graph with the coefficient of $x$.

Solution

1. Graph $g_1$ crosses the y-axis at $3$ and falls $2$ units for every $1$ unit to the right, so its slope is $-2$. Thus $g_1$ matches equation 1, $y = -2x + 3$. 2. Graph $g_2$ crosses the y-axis at $1$ and rises $1$ unit for every $1$ unit to the right, so its slope is $1$. Thus $g_2$ matches equation 3, $y = x + 1$. 3. Graph $g_3$ crosses the y-axis at $-2$ and rises $1$ unit for every $3$ units to the right, so its slope is $\frac{1}{3}$. Thus $g_3$ matches equation 2, $y = \frac{1}{3}x - 2$.

Answer

$g_1$: 1) $y = -2x + 3$; $g_2$: 3) $y = x + 1$; $g_3$: 2) $y = \frac{1}{3}x - 2$

Interpret rate of change and initial value 19

A storage tank contains $10\,\text{gal}$ of water at the start. It is then filled at a constant rate. The graph shows the volume $V$, in gallons, after $t$ minutes. The axes use different scales. a) Find the linear function $V(t)$. b) Find the amount of water in the tank after $6$ minutes. Then check your result on the graph.

Hints

- First check the scale used on each axis. - Choose two points on the line that lie exactly on grid intersections. - The point where the line crosses the vertical axis gives the initial value. - Substitute the requested time into your function.

Solution

1. Two convenient points are $(0, 10)$ and $(4, 20)$. 2. The slope is $m = \frac{20 - 10}{4 - 0} = 2.5$. The y-intercept is $10$, so $V(t) = 2.5t + 10$. 3. $V(6) = 2.5 \cdot 6 + 10 = 25$. The graph also contains the point $(6, 25)$.

Answer

a) $V(t) = 2.5t + 10$ b) $25\,\text{gal}$

Interpret rate of change and initial value 20

Consider the linear equation $x+2y=4$. a) Rewrite the equation in the form $y=mx+b$. b) State the rate of change and the initial value. Interpret the rate by describing how $y$ changes when $x$ increases by $2$, and interpret the initial value as a point on the graph. c) Start at the initial-value point. Increase $x$ by $2$ twice, using the rate of change each time to determine the new $y$-value. List the two resulting points.

Hints

- Isolate $y$ so the coefficient of $x$ and the constant term can be interpreted separately. - Translate the rate into a coordinated horizontal and vertical change rather than only naming the coefficient. - The initial-value point occurs when the input is $0$; use that point as the starting location for part c).

Solution

1. From $x+2y=4$, subtract $x$ to get $2y=-x+4$, then divide by $2$: $y=-\frac{1}{2}x+2$. 2. The rate of change is $-\frac{1}{2}$. Thus, when $x$ increases by $2$, $y$ decreases by $1$. The initial value is $2$, so the initial-value point is $(0,2)$. 3. Starting at $(0,2)$, increasing $x$ by $2$ and decreasing $y$ by $1$ gives $(2,1)$. Repeating the same change gives $(4,0)$.

Answer

a) $y=-\frac{1}{2}x+2$ b) Rate of change: $-\frac{1}{2}$; for every increase of $2$ in $x$, $y$ decreases by $1$. Initial value: $2$, represented by $(0,2)$. c) $(2,1)$ and $(4,0)$

Interpret rate of change and initial value 21

Two lines intersect at $S(2, 1)$. Use the graph to find the slope $m$ and y-intercept $b$ of each line.

Hints

- The intersection point is a convenient point to use when finding each slope. - Compare each line's y-intercept with the shared point $S$.

Solution

1. Line $f$ crosses the y-axis at $-2$. From $(0, -2)$ to $S(2, 1)$, the rise is $3$ and the run is $2$, so $m = \frac{3}{2}$. Thus $f(x) = \frac{3}{2}x - 2$. 2. Line $g$ crosses the y-axis at $4$. From $(0, 4)$ to $S(2, 1)$, the rise is $-3$ and the run is $2$, so $m = -\frac{3}{2}$. Thus $g(x) = -\frac{3}{2}x + 4$.

Answer

Line $f$: $m = \frac{3}{2}$, $b = -2$; line $g$: $m = -\frac{3}{2}$, $b = 4$

Interpret rate of change and initial value 22

Which of graphs a)–d) represents the linear function $f(x) = 0.5x + 1$?

Hints

- Compare the y-intercept in the equation with where each graph crosses the y-axis. - Does a positive slope make a graph rise or fall from left to right? - Use a slope triangle: if you move one unit to the right, how far should the line move vertically? - You can also substitute an x-value into the equation and check whether the resulting point lies on a candidate graph.

Solution

1. The y-intercept is $1$, so the correct graph must cross the y-axis at $1$. This eliminates graph c). 2. The slope is positive, so the line must rise from left to right. This eliminates graph d). 3. A slope of $0.5 = \frac{1}{2}$ means the line rises $1$ unit for every $2$ units to the right. Graph b) has this slope, while graph a) is steeper. 4. Therefore, graph b) represents the function.

Answer

Interpret rate of change and initial value 23

The graph of a linear function $g$ is shown. Which equation represents the graph? A. $g(x)=-1.5x+3$ B. $g(x)=1.5x+3$ C. $g(x)=-\frac{2}{3}x+3$ D. $g(x)=-1.5x-3$

Hints

- First find where the graph crosses the y-axis. Which parameter is this in $y = mx + b$? - Use two points that lie exactly on grid intersections to make a slope triangle. - Check the sign of the slope. Does the line rise or fall from left to right? - Compare the slope you found with the slopes in the answer choices.

Solution

1. The line crosses the y-axis at $(0, 3)$, so the y-intercept is $3$. This eliminates D. 2. The line falls from left to right, so its slope is negative. This eliminates B. 3. Use $(0, 3)$ and $(2, 0)$: $m = \frac{0 - 3}{2 - 0} = -\frac{3}{2} = -1.5$. 4. The equation is $g(x) = -1.5x + 3$, which is choice A.

Answer

A: $g(x) = -1.5x + 3$

Interpret rate of change and initial value 24

The graph shows two linear functions, $f(x) = m_1x + b_1$ and $g(x) = m_2x + b_2$. For each function, determine whether its slope $m$ and y-intercept $b$ are positive or negative.

Hints

- A line that rises from left to right has what sign of slope? - Look at where each line crosses the y-axis. Is that point above or below $0$?

Solution

1. The graph of $f$ rises from left to right, so $m_1 > 0$. It crosses the y-axis below the origin, so $b_1 < 0$. 2. The graph of $g$ falls from left to right, so $m_2 < 0$. It crosses the y-axis above the origin, so $b_2 > 0$.

Answer

For $f$: $m_1 > 0$ and $b_1 < 0$; for $g$: $m_2 < 0$ and $b_2 > 0$.

Interpret rate of change and initial value 25

Which equation represents the line shown in the coordinate plane? A) $y = \frac{1}{2}x + 2$ B) $y = -x + 2$ C) $y = -2x + 2$ D) $y = -\frac{1}{2}x + 2$

Hints

- First identify where the line crosses the y-axis. - Use two grid points to calculate rise over run. - Check whether the line rises or falls from left to right to determine the sign of the slope.

Solution

1. The line crosses the y-axis at $2$, so its y-intercept is $2$. All four choices have this y-intercept. 2. Use $(0, 2)$ and $(4, 0)$: $m = \frac{0 - 2}{4 - 0} = -\frac{1}{2}$. 3. The equation is $y = -\frac{1}{2}x + 2$, which is choice D.

Answer

D) $y = -\frac{1}{2}x + 2$

Interpret rate of change and initial value 26

Water enters a container at a constant rate of $4$ gallons per minute. After $3$ minutes, the container holds $23$ gallons. How much water was there at the start?

Hints

- The current amount includes water added since the start. - Find how much was added during the three minutes, then work backward.

Solution

1. The inflow added $4\cdot3=12$ gallons in three minutes. 2. Subtract from the observed volume: $23-12=11$ gallons at the start.

Answer

$11$ gallons

Interpret rate of change and initial value 27

Two rain barrels hold water. The graph shows the amount $A(t)$, in liters, in the first barrel after $t$ minutes. The second barrel holds $B(t)=9+t$ liters after $t$ minutes. a) Which barrel starts with more water? Give both starting amounts. b) Which barrel is filling, and which is draining? Give the rate for each, in liters per minute.

Hints

- Compare both functions when time equals zero. - How does the graphed amount change over equal time intervals? - Compare that change with the amount added by the second barrel's rule.

Solution

1. The graph shows $A(0)=12$ liters; the rule gives $B(0)=9$ liters. The first barrel starts with more. 2. The graph drops from $12$ to $8$ liters in $2$ minutes, so $A$ drains at $2$ liters per minute. From the equation, $B$ fills at $1$ liter per minute.

Answer

a) The first barrel: $12$ liters compared with $9$ liters b) The first drains at $2$ liters per minute; the second fills at $1$ liter per minute.

Interpret rate of change and initial value 28

Anna and Ben start a bike ride at the same time and place. Anna rides at a constant speed, while Ben changes speed during the first hour. The table shows their distances from the starting point. <table><tr><td>Time (min)</td><td>$0$</td><td>$15$</td><td>$30$</td><td>$45$</td><td>$60$</td></tr><tr><td>Anna's distance (mi)</td><td>$0$</td><td>$3$</td><td>$6$</td><td>$9$</td><td>$12$</td></tr><tr><td>Ben's distance (mi)</td><td>$0$</td><td>$4$</td><td>$7$</td><td>$9$</td><td>$10$</td></tr></table> Assume the distance changes linearly between consecutive measurements. a) Describe both riders' distance-time graphs, with time on the x-axis and distance on the y-axis. b) Who is ahead after $20$ minutes, and by how much? c) Find Anna's average speed in miles per hour. d) Compare Ben's ride with Anna's ride. Give a possible reason for Ben's changes in speed.

Hints

- Compare the slopes implied by consecutive table values. - Use the linear change between the two measurements surrounding $20$ minutes. - One hour is $60$ minutes. - A less steep distance-time graph represents a lower speed.

Solution

1. Anna's points $(0, 0)$, $(15, 3)$, $(30, 6)$, $(45, 9)$, and $(60, 12)$ lie on one line. Ben's points $(0, 0)$, $(15, 4)$, $(30, 7)$, $(45, 9)$, and $(60, 10)$ form a piecewise-linear graph with segments that become less steep. 2. Anna travels $3$ miles every $15$ minutes. After $20$ minutes, she has traveled $3+\frac{5}{15}(3)=4$ miles. Between minutes $15$ and $30$, Ben travels $3$ miles, so after $20$ minutes he has traveled $4+\frac{5}{15}(3)=5$ miles. Ben is ahead by $1$ mile. 3. Anna travels $12$ miles in $1$ hour, so her average speed is $12\,\text{mph}$. 4. Anna's graph is a line with constant slope. Ben's line segments become less steep, so his speed decreases. Fatigue, a headwind, or an uphill section could explain the change.

Answer

a) Anna's graph is a line; Ben's graph is a piecewise-linear graph whose segments become less steep. b) Ben is ahead by $1$ mile. c) $12\,\text{mph}$ d) Ben starts faster but slows down, while Anna maintains a constant speed. A possible cause is fatigue, wind, or an uphill section.

Interpret rate of change and initial value 29

A new smartphone and an older model are charged under the same conditions. Their battery levels are recorded every $10$ minutes. <table><tr><td>Time (min)</td><td>$0$</td><td>$10$</td><td>$20$</td><td>$30$</td><td>$40$</td><td>$50$</td><td>$60$</td></tr><tr><td>New model (%)</td><td>$5$</td><td>$25$</td><td>$45$</td><td>$60$</td><td>$72$</td><td>$82$</td><td>$90$</td></tr><tr><td>Older model (%)</td><td>$5$</td><td>$15$</td><td>$25$</td><td>$35$</td><td>$45$</td><td>$55$</td><td>$65$</td></tr></table> a) For the new model, find the average charging rate from $0$ to $20$ minutes and from $40$ to $60$ minutes, in percentage points per minute. b) Compare the charging patterns of the two phones. c) If the older model's constant pattern continues, when will it reach $100\%$? d) Why might the new model's charging rate decrease near a full charge?

Hints

- Divide the change in battery level by the elapsed time. - Check whether the older model gains the same amount in every interval. - Find how many percentage points the older model still needs after $60$ minutes.

Solution

1. From $0$ to $20$ minutes, the new model increases by $45-5=40$ percentage points, so its average rate is $40\div20=2$ percentage points per minute. From $40$ to $60$ minutes, it increases by $90-72=18$ percentage points, so its average rate is $18\div20=0.9$ percentage point per minute. 2. The older model gains a constant $10$ percentage points every $10$ minutes. The new model charges faster at first, but its charging rate decreases as its battery level rises. 3. The older model gains $1$ percentage point per minute. At $60$ minutes, it needs $35$ more percentage points, so it needs $35$ more minutes. It reaches $100\%$ after $95$ minutes. 4. Charging systems often reduce current near a full charge to limit heat and battery stress.

Answer

a) $0$–$20$ min: $2$ percentage points per minute; $40$–$60$ min: $0.9$ percentage point per minute b) The older model charges at a constant rate; the new model charges quickly at first and then more slowly. c) After $95$ minutes d) The charging current may be reduced to limit heat and protect the battery.

Interpret rate of change and initial value 30

A rectangular water tank is being filled at a constant rate. The water height $h$, in centimeters, after $t$ minutes is modeled by $h=3t+10$. a) Someone claims, “The tank was empty when filling began.” Use the equation to evaluate this claim. b) How many centimeters does the water level rise each minute? c) After how many minutes will the water reach a height of $1.60\,\text{m}$?

Hints

- What does substituting $t=0$ tell you? - Compare the units used in the equation with the unit in part c). - Which number in the equation represents the change in height for each additional minute?

Solution

1. At $t=0$, the model gives $h=3\cdot0+10=10$. The tank already had $10\,\text{cm}$ of water, so the claim is false. 2. The coefficient of $t$ is $3$, so the water level rises $3\,\text{cm}$ per minute. 3. Convert $1.60\,\text{m}$ to $160\,\text{cm}$. Solve $160=3t+10$: $150=3t$, so $t=50$. 4. The water reaches the stated height after $50$ minutes.

Answer

a) The claim is false. At $t=0$, the water height is $10\,\text{cm}$. b) $3\,\text{cm}$ per minute c) $50$ minutes

Interpret rate of change and initial value 31

A simplified hypothetical data set gives the amount of plastic waste produced per person each year in a country. <table><thead><tr><th>Year</th><th>2000</th><th>2004</th><th>2008</th><th>2012</th><th>2016</th><th>2020</th></tr></thead><tbody><tr><td>Waste per person (kg)</td><td>$32$</td><td>$35$</td><td>$39$</td><td>$42$</td><td>$44$</td><td>$45$</td></tr></tbody></table> a) Find the average yearly increase from 2000 to 2012. b) Use linear interpolation to estimate the amount for 2018. c) A hypothetical environmental organization predicts that the amount will exceed $60\,\text{kg}$ per person by 2030. Does this prediction seem realistic based on the data from 2012 to 2020? Explain.

Hints

- Divide the total change by the number of years. - A year halfway between two table entries has the midpoint of their values under a linear model. - Compare the recent rate with the rate required by the prediction.

Solution

1. From 2000 to 2012, the total increase is $42-32=10\,\text{kg}$ over $12$ years. The average yearly increase is $10\div12\approx0.83\,\text{kg}$ per year. 2. The year 2018 is halfway between 2016 and 2020, so linear interpolation gives $(44+45)\div2=44.5\,\text{kg}$. 3. From 2012 to 2020, the amount increased by $45-42=3\,\text{kg}$ over $8$ years, or $3\div8=0.375\,\text{kg}$ per year. 4. To rise from $45\,\text{kg}$ in 2020 to more than $60\,\text{kg}$ in 2030 would require an average increase greater than $15\div10=1.5\,\text{kg}$ per year. That is much faster than the recent rate, so the prediction does not appear realistic from these data.

Answer

a) $\approx0.83\,\text{kg}$ per year b) $44.5\,\text{kg}$ c) The prediction does not appear realistic. The recent rate was $0.375\,\text{kg}$ per year, while exceeding $60\,\text{kg}$ would require more than $1.5\,\text{kg}$ per year.

Interpret rate of change and initial value 32

A line $h$ passes through the y-intercept $(0, 4)$ and the x-intercept $(-2, 0)$. a) Use the two intercepts to find the slope $m$ of $h$. b) Write the equation of $h$. c) A second line $k$ is parallel to $h$ and has y-intercept $(0, -2)$. Write the equation of $k$ and identify its x-intercept.

Hints

- Compare the vertical and horizontal changes between the two intercepts. - Use the slope and y-intercept in $y = mx + b$. - What must be true about the slopes of parallel lines? - Set $k(x) = 0$ to find its x-intercept.

Solution

1. From $(-2, 0)$ to $(0, 4)$, the rise is $4$ and the run is $2$, so $m = \frac{4}{2} = 2$. 2. Line $h$ has slope $2$ and y-intercept $4$, so $h(x) = 2x + 4$. 3. A parallel line has the same slope, so $k(x) = 2x - 2$. Its x-intercept satisfies $0 = 2x - 2$, giving $x = 1$. Therefore, the x-intercept is $(1, 0)$.

Answer

a) $m = 2$ b) $h(x) = 2x + 4$ c) $k(x) = 2x - 2$, with x-intercept $(1, 0)$.

Interpret rate of change and initial value 33

A candle's height is modeled by $h(t)=8-t$, where $h(t)$ is the height in inches after $t$ hours. a) State the initial value and the rate of change, including units, and explain what each means for the candle. b) Find $h(3)$ and $h(5.5)$. c) When is the candle exactly $3\,\text{in.}$ tall? d) Find $h(10)$. Interpret the result and state a reasonable domain for the model.

Hints

- In a linear rule, locate the output at input $0$ and the amount the output changes when the input increases by $1$. - Keep the sign of the rate tied to whether the candle is growing or shrinking. - The physical domain ends when the modeled height reaches $0$.

Solution

1. The initial value is $8\,\text{in.}$, so the candle is $8\,\text{in.}$ tall at $t=0$. The rate of change is $-1\,\frac{\text{in.}}{\text{h}}$, meaning the height decreases by $1\,\text{in.}$ each hour. 2. $h(3)=8-3=5$, and $h(5.5)=8-5.5=2.5$. 3. Solve $3=8-t$. This gives $t=5$, so the candle is $3\,\text{in.}$ tall after $5$ hours. 4. $h(10)=8-10=-2$. A negative candle height is not physically meaningful. The model is reasonable from lighting until the candle reaches height $0$, so $0\le t\le8$.

Answer

a) Initial value: $8\,\text{in.}$; rate of change: $-1\,\frac{\text{in.}}{\text{h}}$, meaning the candle loses $1\,\text{in.}$ of height each hour. b) $h(3)=5\,\text{in.}$ and $h(5.5)=2.5\,\text{in.}$ c) $5$ hours d) $h(10)=-2$, which is not physically meaningful; a reasonable domain is $0\le t\le8$.

Interpret rate of change and initial value 34

A rain barrel fills at a constant rate. Its volume in gallons after $t$ minutes is $V(t)=0.5t+5$. a) What do $5$ and $0.5$ mean here? b) How much water is in the barrel after $12$ minutes? c) The barrel holds at most $25$ gallons. Give a reasonable time domain for the model. d) Describe the graph on that domain. How would it change if the barrel started empty but filled at the same rate?

Hints

- Use the output at $t=0$ for the initial amount. - The coefficient of time describes change in gallons per minute. - Find the first time the full capacity is reached for the reasonable domain. - Both graphs have the same filling rate but different starting amounts.

Solution

1. The initial volume is $5$ gallons and the filling rate is $0.5$ gallon per minute. 2. $V(12)=0.5\cdot12+5=11$ gallons. 3. The barrel reaches $25$ gallons when $0.5t+5=25$, so $t=40$ and the reasonable domain is $0\leq t\leq40$. 4. The original graph is a line segment from $(0,5)$ to $(40,25)$. Starting empty keeps the rate but changes the segment to run from $(0,0)$ to $(50,25)$.

Answer

a) Initial volume: $5$ gallons; filling rate: $0.5$ gallon per minute. b) $11$ gallons c) $0\leq t\leq40$ minutes d) Original graph: segment from $(0,5)$ to $(40,25)$. Empty-start graph: same slope, from $(0,0)$ to $(50,25)$.

Interpret rate of change and initial value 35

A hiker descends from a mountain lodge. The hiker's elevation in feet is modeled by $h(t)=5000-800t$, where $t$ is the number of hours since departure. a) State the rate of change and the initial value of $h$. Interpret each value in this situation, including the sign and units of the rate. b) Find the hiker's elevation after $2.5$ hours. c) When does the hiker reach an elevation of $2200\,\text{ft}$? When does the hiker reach the valley at elevation $0\,\text{ft}$? State the reasonable domain and range of $h$ for this trip. d) A second hiker starts at $4000\,\text{ft}$ and descends at $600\,\text{ft}$ per hour. Write a linear model $g(t)$ for the second hiker's elevation. State its rate of change and initial value, then determine which hiker reaches the valley first.

Hints

- Compare each linear model with the structure $y=mx+b$, and keep track of the units attached to the coefficient of time. - Ask what the output is at $t=0$ and what that output represents in the trip. - For a requested elevation, set the model's output equal to that elevation and solve for time. - For the second hiker, combine the starting elevation with the signed hourly change before comparing valley-arrival times.

Solution

1. In $h(t)=5000-800t$, the rate of change is $-800\,\text{ft/h}$. The negative sign means the hiker's elevation decreases by $800\,\text{ft}$ each hour. The initial value is $h(0)=5000\,\text{ft}$, the elevation at departure. 2. $h(2.5)=5000-800(2.5)=3000$, so the elevation is $3000\,\text{ft}$. 3. For $2200\,\text{ft}$, solve $2200=5000-800t$. This gives $t=3.5$ hours. For the valley, solve $0=5000-800t$, giving $t=6.25$ hours. Therefore the reasonable domain is $[0,6.25]$ hours and the range is $[0,5000]$ feet. 4. The second hiker's model is $g(t)=4000-600t$. Its rate of change is $-600\,\text{ft/h}$, and its initial value is $4000\,\text{ft}$. Solving $g(t)=0$ gives $t=\frac{20}{3}\approx6.67$ hours. Since $6.25<6.67$, the first hiker reaches the valley first.

Answer

a) Rate of change: $-800\,\text{ft/h}$, meaning the elevation decreases by $800\,\text{ft}$ each hour. Initial value: $5000\,\text{ft}$, the departure elevation. b) $3000\,\text{ft}$ c) $3.5$ hours to reach $2200\,\text{ft}$; $6.25$ hours to reach the valley; domain $[0,6.25]$, range $[0,5000]$. d) $g(t)=4000-600t$; rate $-600\,\text{ft/h}$; initial value $4000\,\text{ft}$. The first hiker reaches the valley first.

Interpret rate of change and initial value 36

The linear functions are $f(x) = \frac{2}{3}x + 1$ and $g(x) = 1.5x + 1$. a) What point do the two graphs have in common? b) For each function, consider a slope triangle with a horizontal change of $\Delta x = 6$. What is the corresponding vertical change $\Delta y$? c) Which line is steeper? Explain your reasoning.

Hints

- Where can you read the y-intercept directly from each equation? - How does slope connect a horizontal change to a vertical change? - What does the size of the slope tell you about a line's steepness?

Solution

1. Both functions have y-intercept $1$. When $x = 0$, both have output $1$, so the common point is $(0, 1)$. 2. For $f$, $\Delta y = m\Delta x = \frac{2}{3} \cdot 6 = 4$. 3. For $g$, $\Delta y = 1.5 \cdot 6 = 9$. 4. Since $1.5 > \frac{2}{3}$, line $g$ is steeper. The larger vertical change for the same horizontal change confirms this.

Answer

a) $(0, 1)$ b) For $f$, $\Delta y = 4$; for $g$, $\Delta y = 9$. c) $g$ is steeper because its slope is larger.

Interpret rate of change and initial value 37

Consider the four linear functions: $g_1(x) = 1.5x + 2$ $g_2(x) = 1.5x - 4$ $g_3(x) = -x + 2$ $g_4(x) = 3x + 2$. Use the slope $m$ and y-intercept $b$ to answer each question. a) Which graphs are parallel? b) Which graphs have the same y-intercept? Give the point. c) Which graph increases most steeply?

Hints

- What must be true about the slopes of parallel lines? - Which parameter gives the y-intercept? - Compare the positive slopes to determine which line rises fastest.

Solution

1. Parallel lines have the same slope. Both $g_1$ and $g_2$ have slope $1.5$, so they are parallel. 2. The y-intercept is determined by $b$. Functions $g_1$, $g_3$, and $g_4$ all have $b = 2$, so they pass through $(0, 2)$. 3. Among the positive slopes, $3$ is the greatest. Therefore, $g_4$ increases most steeply.

Answer

a) $g_1$ and $g_2$ b) $g_1$, $g_3$, and $g_4$; the common y-intercept is $(0, 2)$. c) $g_4$

Interpret rate of change and initial value 38

Consider the two lines $k$ and $l$: $k: y = \frac{2}{3}x + 1$ $l: y = 0.6x + 1$ Bram draws a slope triangle for each line. For line $k$, the horizontal change is $3$ units. For line $l$, the horizontal change is $5$ units. a) Find the vertical change for each slope triangle. b) Which line is steeper? Justify your answer by comparing the slopes.

Hints

- Relate slope to horizontal and vertical change for each triangle. - Compare the two slopes in a common numerical form. - Distinguish the height of one chosen slope triangle from the line's slope itself.

Solution

1. For line $k$, $\Delta y = \frac{2}{3} \cdot 3 = 2$. 2. For line $l$, $\Delta y = 0.6 \cdot 5 = 3$. 3. The slopes are $\frac{2}{3}$ and $0.6$. Since $\frac{2}{3} \approx 0.667 > 0.6$, line $k$ is steeper. The larger height of the second slope triangle does not make $l$ steeper because the two triangles have different horizontal widths.

Answer

a) Line $k$: $2\,\text{units}$; line $l$: $3\,\text{units}$ b) Line $k$ is steeper because $\frac{2}{3} > 0.6$.

Interpret rate of change and initial value 39

Consider the family of lines described by $y = -1.5x + b$. a) How does the graph change as $b$ decreases? Describe the shift. b) Find the value of $b$ that makes the x-intercept $6$. c) Explain why every line in this family intersects $h: y = 2x - 4$ at exactly one point, no matter what value $b$ has.

Hints

- What does the constant term in slope-intercept form tell you about the graph? - What is the y-coordinate of an x-intercept? - When are two lines parallel? - What do different slopes imply about two lines?

Solution

1. The parameter $b$ is the y-intercept. Decreasing $b$ shifts the line downward without changing its slope. 2. An x-intercept of $6$ means $y = 0$ when $x = 6$. Thus $0 = -1.5 \cdot 6 + b$, so $b = 9$. 3. Every line in the family has slope $-1.5$, while $h$ has slope $2$. Because the slopes are different, the lines are not parallel, so each line in the family intersects $h$ exactly once.

Answer

a) The line shifts downward as $b$ decreases. b) $b = 9$ c) The slopes are different, $-1.5 \neq 2$, so the lines intersect exactly once.

Interpret rate of change and initial value 40

Consider the family of functions $f(x) = mx - 2$. a) What point is on every graph in this family? Give its coordinates. b) For which values of $m$ does the line decrease from left to right? c) Compare the graphs for $m = 1$ and $m = 2$. Which graph is steeper? Briefly explain. d) Write an equation for a line $g$ that is parallel to the graph with $m = 0.5$ and has y-intercept $3$.

Hints

- What does the constant term tell you about where each graph crosses the y-axis? - How does the sign of the slope affect whether a line increases or decreases? - For two positive slopes, how does the larger slope affect steepness? - What must be the same for two distinct lines to be parallel?

Solution

1. Every graph has y-intercept $-2$, so every graph passes through $(0, -2)$. 2. A line decreases from left to right when its slope is negative, so $m < 0$. 3. The line with $m = 2$ is steeper than the line with $m = 1$ because $2 > 1$. 4. A parallel line must have slope $0.5$. With y-intercept $3$, its equation is $g(x) = 0.5x + 3$.

Answer

a) $(0, -2)$ b) $m < 0$ c) The graph with $m = 2$ is steeper. d) $g(x) = 0.5x + 3$

Interpret rate of change and initial value 41

A line has slope $m = -\frac{2}{3}$. Evaluate each student's proposed slope triangle. a) Sarah says, “I move 3 units right and 2 units up.” b) Tom says, “I move $1.5$ units left and 1 unit up.” Use calculations to determine who is correct.

Hints

- Pay close attention to the sign of the given slope. - What signs do $\Delta x$ and $\Delta y$ have for each move? - How can you compare the decimal ratio with the fraction in the given slope?

Solution

1. Sarah's move has $\Delta x = 3$ and $\Delta y = 2$, so the slope would be $\frac{2}{3}$. This is positive, so Sarah is not correct. 2. Tom's move has $\Delta x = -1.5$ and $\Delta y = 1$, so the slope is $\frac{1}{-1.5} = -\frac{2}{3}$. Tom is correct.

Answer

Tom is correct. Sarah's move gives slope $\frac{2}{3}$, while Tom's move gives slope $-\frac{2}{3}$.

Interpret rate of change and initial value 42

Three descriptions of slope are given for a linear function: (1) “Move 5 units right and 2 units down.” (2) “Move $2.5$ units left and 1 unit up.” (3) “The slope is $m = -2.5$.” Determine which descriptions represent the same slope. Then, using the slope from description (1), find the change in $y$ when $x$ increases by 10 units.

Hints

- Find the numerical slope represented by each description. - How do moving left or down affect the signs of $\Delta x$ and $\Delta y$? - How are slope, $\Delta x$, and $\Delta y$ related?

Solution

1. Description (1) gives $m = \frac{-2}{5} = -0.4$. 2. Description (2) gives $m = \frac{1}{-2.5} = -0.4$. 3. Description (3) gives $m = -2.5$, so it is different from the first two. 4. For $m = -0.4$ and $\Delta x = 10$, $\Delta y = -0.4 \cdot 10 = -4$. Thus, $y$ decreases by 4 units.

Answer

Descriptions (1) and (2) represent the same slope, $m = -0.4$. When $x$ increases by 10 units, $\Delta y = -4$, so $y$ decreases by 4 units.

Interpret rate of change and initial value 43

Two linear functions $f$ and $g$ have slopes $m_f = 1.25$ and $m_g = 0.8$. a) For each function, find the vertical change $\Delta y$ when the horizontal change is $\Delta x = 4$. b) For function $f$, describe a slope triangle that moves to the left and has whole-number side lengths.

Hints

- Use the relationship $\Delta y = m\Delta x$. - How can you rewrite $1.25$ as a fraction to choose whole-number changes? - For a positive slope, what direction must $y$ change when you move left?

Solution

1. For $f$, $\Delta y = 1.25 \cdot 4 = 5$. 2. For $g$, $\Delta y = 0.8 \cdot 4 = 3.2$. 3. Since $1.25 = \frac{5}{4}$, choose $\Delta x = -4$. Then $\Delta y = 1.25 \cdot (-4) = -5$. One suitable triangle moves 4 units left and 5 units down.

Answer

a) For $f$, $\Delta y = 5$. For $g$, $\Delta y = 3.2$. b) One possible slope triangle is 4 units left and 5 units down.

Interpret rate of change and initial value 44

Consider the functions $f(x) = mx + 3$ and $g(x) = -mx + 3$, where $m > 0$. 1. Without calculating, explain why the two graphs always intersect at $(0, 3)$. 2. How does increasing $m$ affect the two lines? 3. The graph of $f$ passes through $P(2, 7)$. Find $m$, write the equation of $g$, and give one additional point $Q$ on $g$ that is not on the y-axis.

Hints

- What does the constant term in slope-intercept form tell you about a graph? - Think about the vertical change for a horizontal change of 1 as $m$ gets larger. - Substitute the coordinates of $P$ into $f(x) = mx + 3$ to find $m$.

Solution

1. Both functions have y-intercept $3$, so both graphs pass through $(0, 3)$. 2. As $m$ increases, the graph of $f$ rises more steeply and the graph of $g$, whose slope is $-m$, falls more steeply. 3. Since $P(2, 7)$ lies on $f$, $7 = 2m + 3$. Thus, $m = 2$, so $g(x) = -2x + 3$. For example, $g(1) = 1$, so $Q(1, 1)$ is on the graph.

Answer

1. Both graphs have y-intercept $3$, so they share $(0, 3)$. 2. $f$ becomes steeper increasing, while $g$ becomes steeper decreasing. 3. $m = 2$; $g(x) = -2x + 3$; one possible point is $Q(1, 1)$.

Interpret rate of change and initial value 45

Consider the functions 1) $f(x)=1.5x$ 2) $g(x)=1.5x+4$ 3) $h(x)=3$ a) Which function describes a proportional relationship? Justify your choice using its initial value. b) State the rate of change and initial value of each function. c) Test the doubling rule for $g(x)$ by finding $g(2)$ and $g(4)$ and comparing them.

Hints

- Evaluate each function at $x=0$ to find its initial value. - Identify the coefficient that gives the rate of change. - In a proportional relationship, doubling the input should double the output.

Solution

1. Only $f(x)=1.5x$ is proportional because $f(0)=0$, so its graph passes through $(0, 0)$. 2. For $f$, the rate of change is $1.5$ and the initial value is $0$. For $g$, the rate of change is $1.5$ and the initial value is $4$. For $h$, the rate of change is $0$ and the initial value is $3$. 3. $g(2)=1.5\cdot 2+4=7$, and $g(4)=1.5\cdot 4+4=10$. 4. Since $10\ne 2\cdot 7$, doubling the input does not double the output for $g$.

Answer

a) Only $f(x)$, because its initial value is $0$. b) $f$: rate $1.5$, initial value $0$. $g$: rate $1.5$, initial value $4$. $h$: rate $0$, initial value $3$. c) $g(2)=7$ and $g(4)=10$, so the doubling rule does not hold.

Interpret rate of change and initial value 46

The linear function $f$ is given by $f(x) = 1.5x - 3$. a) Find the x-intercept and y-intercept of the graph. b) Martín says, “From any point on the line, if I move $2\,\text{units}$ right, I must move exactly $3\,\text{units}$ up to land on the line again.” Check whether this statement matches the slope. c) How does the graph change if the constant term $-3$ is replaced by $+1$? Describe the change precisely.

Hints

- Ask which coordinate is zero when a graph crosses each coordinate axis. - Relate a horizontal and vertical step to the slope coefficient. - Compare the old and new constant terms while keeping the slope fixed.

Solution

1. At the y-intercept, $x = 0$, so the point is $(0, -3)$. At the x-intercept, $0 = 1.5x - 3$, so $x = 2$ and the point is $(2, 0)$. 2. A move of $2\,\text{units}$ right and $3\,\text{units}$ up gives slope $\frac{3}{2} = 1.5$, so Martín's statement is correct. 3. Changing the constant term from $-3$ to $1$ increases the y-intercept by $4$ while leaving the slope unchanged. The entire line shifts up $4\,\text{units}$.

Answer

a) x-intercept: $(2, 0)$; y-intercept: $(0, -3)$ b) The statement is correct because $\frac{3}{2} = 1.5$. c) The line shifts up $4\,\text{units}$.

Interpret rate of change and initial value 47

A rain barrel has a narrow lower section and a wider upper section. During steady rain, the water height is recorded. <table><tr><td>Time $t$ (min)</td><td>$0$</td><td>$10$</td><td>$20$</td><td>$30$</td><td>$40$</td><td>$50$</td></tr><tr><td>Water height $h$ (in.)</td><td>$0$</td><td>$5$</td><td>$10$</td><td>$12.5$</td><td>$15$</td><td>$17.5$</td></tr></table> a) Determine whether the entire relation from time to water height is linear. b) Find the average rates of change, in inches per minute, on $[0, 20]$ and $[20, 50]$. c) Interpret the two rates in terms of the barrel's shape.

Hints

- Calculate slopes for different parts of the table. - With a constant inflow, a wider section causes the water level to rise more slowly. - Compare the two rates directly.

Solution

1. From $0$ to $10$ minutes, the rate is $\frac{5-0}{10-0}=0.5$ inch per minute. From $20$ to $30$ minutes, it is $\frac{12.5-10}{30-20}=0.25$ inch per minute. The rate is not constant over the entire table, so the relation is not linear. 2. On $[0, 20]$, the rate is $\frac{10-0}{20-0}=0.5$ inch per minute. On $[20, 50]$, the rate is $\frac{17.5-10}{50-20}=\frac{7.5}{30}=0.25$ inch per minute. 3. The water rises twice as fast in the lower section. Because the rainfall rate is constant, the smaller rate above $10$ inches indicates that the barrel becomes wider there.

Answer

a) No b) $[0, 20]$: $0.5$ inch per minute; $[20, 50]$: $0.25$ inch per minute c) The barrel becomes wider above a height of $10$ inches.

Interpret rate of change and initial value 48

A hiking group records its distance during a trail hike. <table> <tr><td>Time $t$, in hours</td><td>$0$</td><td>$1$</td><td>$2$</td><td>$2.5$</td><td>$3$</td><td>$4.5$</td></tr> <tr><td>Distance $s$, in miles</td><td>$0$</td><td>$3$</td><td>$6$</td><td>$6$</td><td>$8$</td><td>$14$</td></tr> </table> a) Find the average speed for each interval between consecutive measurements. b) Can one proportional function describe the entire hike? Justify your answer. c) What most likely happened from $t=2$ to $t=2.5$? d) Compare the group's walking speed before and after that interval.

Hints

- Speed is the slope over each time interval. - What does a constant distance over time mean? - Check whether one rate describes every interval.

Solution

1. The interval speeds are $(3-0)\div(1-0)=3$, $(6-3)\div(2-1)=3$, $(6-6)\div(2.5-2)=0$, $(8-6)\div(3-2.5)=4$, and $(14-8)\div(4.5-3)=4$ miles per hour. 2. One proportional function cannot describe the entire hike because the rate of change is not constant. 3. The unchanged distance from $2$ to $2.5$ hours indicates a $30$-minute break. 4. Before the break, the group walked at $3$ miles per hour; after the break, it walked at $4$ miles per hour.

Answer

a) $3$, $3$, $0$, $4$, and $4$ miles per hour. b) No. The rate of change is not constant. c) The group took a $30$-minute break. d) The group walked faster after the break: $4$ miles per hour instead of $3$.

Interpret rate of change and initial value 49

Two water tanks are being filled. Their water heights $h$, in centimeters, after $t$ minutes are modeled by Tank A: $h_A(t) = 1.5t + 20$ Tank B: $h_B(t) = 2t + 10$ a) What is the initial water height in each tank? b) How many centimeters per minute does the water level rise in each tank? c) Which tank has the higher water level after $15$ minutes? Show your calculation. d) Find the time when the water levels are equal.

Hints

- Which part of a linear function gives the starting value? - What does the slope represent in this context? - Substitute the given time into each function to compare the water heights. - To find when the heights are equal, set the two function expressions equal.

Solution

1. The initial heights are the values at $t = 0$: Tank A starts at $20\,\text{cm}$, and Tank B starts at $10\,\text{cm}$. 2. The rates of change are the slopes: Tank A rises at $1.5\,\text{cm/min}$, and Tank B rises at $2\,\text{cm/min}$. 3. After $15$ minutes, $h_A(15) = 1.5 \cdot 15 + 20 = 42.5\,\text{cm}$, while $h_B(15) = 2 \cdot 15 + 10 = 40\,\text{cm}$. Tank A is higher. 4. Set the heights equal: $1.5t + 20 = 2t + 10$. Then $10 = 0.5t$, so $t = 20$ minutes.

Answer

a) Tank A: $20\,\text{cm}$; Tank B: $10\,\text{cm}$ b) Tank A: $1.5\,\text{cm/min}$; Tank B: $2\,\text{cm/min}$ c) Tank A, with $42.5\,\text{cm}$ compared with $40\,\text{cm}$ d) $20$ minutes

Interpret rate of change and initial value 50

A smartphone battery is charging. The model $L(t)=0.8t+15$ gives the battery level $L$, in percent, after $t$ minutes. a) Interpret $15$ and $0.8$ in context. b) What battery level does the model predict after half an hour? c) How long does the model predict it will take to charge from $15\%$ to $100\%$? d) Describe the graph of the function for $0\le t\le60$.

Hints

- Evaluate the function at $t=0$. - Convert half an hour to minutes. - Set the function equal to $100$ and solve. - Two points determine the line segment.

Solution

1. The value $15$ is the initial battery level, $15\%$. The coefficient $0.8$ is the charging rate, $0.8$ percentage point per minute. 2. Half an hour is $30$ minutes. $L(30)=0.8(30)+15=39$, so the predicted battery level is $39\%$. 3. Solve $0.8t+15=100$. Then $0.8t=85$, so $t=106.25$ minutes. 4. Over the first $60$ minutes, the graph is the line segment from $(0, 15)$ to $(60, 63)$.

Answer

a) Initial level: $15\%$; charging rate: $0.8$ percentage point per minute b) $39\%$ c) $106.25$ minutes d) The line segment through $(0, 15)$ and $(60, 63)$

Interpret rate of change and initial value 51

Consider the two linear functions $f(x) = \frac{2}{3}x + 4$ and $g(x) = 0.6x + 4$. a) Which line increases more steeply? Justify your answer by comparing the slopes. b) Find the y-intercept of each line. What do you notice? c) A new line $h$ has twice the slope of $f$ and the same y-intercept as $f$. Write an equation for $h$.

Hints

- Rewrite the slopes in the same form before comparing them. - Where is the y-intercept visible in slope-intercept form? - What happens to the slope when it is doubled?

Solution

1. The slopes are $\frac{2}{3}$ and $0.6$. Since $\frac{2}{3} \approx 0.667 > 0.6$, $f$ is steeper. 2. Both functions have y-intercept $4$, so both graphs pass through $(0, 4)$. 3. Twice the slope of $f$ is $2 \cdot \frac{2}{3} = \frac{4}{3}$. Keeping the y-intercept $4$ gives $h(x) = \frac{4}{3}x + 4$.

Answer

a) $f$ is steeper because $\frac{2}{3} > 0.6$. b) Both y-intercepts are $(0, 4)$. c) $h(x) = \frac{4}{3}x + 4$

Interpret rate of change and initial value 52

Consider the equation $4x + 8y = 16$. a) Find the intercepts of the corresponding line with the coordinate axes. b) Find the slope of the line. c) Explain how $y$ changes when $x$ increases by 3 units.

Hints

- What is always true about the x-coordinate of a point on the y-axis? - In a linear function, how does $y$ change when $x$ changes? Which parameter describes that relationship?

Solution

1. For the y-intercept, set $x = 0$: $8y = 16$, so $y = 2$. The y-intercept is $(0, 2)$. 2. For the x-intercept, set $y = 0$: $4x = 16$, so $x = 4$. The x-intercept is $(4, 0)$. 3. Solve for $y$: $8y = -4x + 16$, so $y = -0.5x + 2$. Thus, the slope is $-0.5$. 4. A slope of $-0.5$ means $\Delta y = -0.5\Delta x$. If $\Delta x = 3$, then $\Delta y = -1.5$, so $y$ decreases by $1.5$ units.

Answer

a) x-intercept: $(4, 0)$; y-intercept: $(0, 2)$ b) $m = -0.5$ c) $y$ decreases by $1.5$ units.

Interpret rate of change and initial value 53

Consider the linear equation $0.5x+2y=4$. a) Rewrite it in the form $y=mx+b$. b) State the rate of change and initial value. c) Starting from the initial value at $x=0$, use the rate of change to find the output when $x=4$. d) Determine whether $(-4,3)$ is an input-output pair of the function.

Hints

- Rewrite the equation so the rate and initial value can be read from $y=mx+b$. - A rate of $-\frac{1}{4}$ tells how the output changes for each one-unit increase in input. - For part c), use the initial value as the starting output instead of solving the original equation from scratch.

Solution

1. Subtract $0.5x$: $2y=4-0.5x$. Divide by $2$: $y=-0.25x+2$. 2. The rate of change is $-0.25=-\frac{1}{4}$, and the initial value is $2$. 3. Increasing $x$ from $0$ to $4$ changes the output by $-\frac{1}{4}\cdot4=-1$. Starting from $2$, the output is $1$, so $(4,1)$ is on the graph. 4. At $x=-4$, $y=-0.25(-4)+2=3$, so $(-4,3)$ is an input-output pair.

Answer

a) $y=-0.25x+2$ b) Rate of change: $-0.25$; initial value: $2$ c) $1$ d) Yes.

Interpret rate of change and initial value 54

Consider the line $2x - 5y = 10$. a) Find the x-intercept and y-intercept. b) Rewrite the equation in the form $y = mx + b$ and state the slope $m$.

Hints

- Set one coordinate equal to zero to find each intercept. - Isolate $y$ to rewrite the equation in slope-intercept form.

Solution

1. For the x-intercept, set $y = 0$: $2x = 10$, so the point is $(5, 0)$. 2. For the y-intercept, set $x = 0$: $-5y = 10$, so the point is $(0, -2)$. 3. Solve for $y$: $-5y = -2x + 10$, so $y = 0.4x - 2$. Therefore, $m = 0.4$.

Answer

a) x-intercept: $(5, 0)$; y-intercept: $(0, -2)$ b) $y = 0.4x - 2$, with slope $m = 0.4$

Interpret rate of change and initial value 55

An empty rain barrel fills at a constant rate for $20$ minutes and receives $12$ gallons. The rain then stops for $10$ minutes. Over the next $15$ minutes, $6$ gallons are removed at a constant rate. a) Describe the volume-versus-time graph for the first $45$ minutes by listing its vertices. b) Find the filling rate in gallons per minute. c) At what times does the barrel contain exactly $9$ gallons?

Hints

- Mark the time and volume at the end of each of the three phases. - A constant rate produces a straight line segment on a volume-time graph. - Check for the target volume during both the increase and the decrease.

Solution

1. The graph consists of straight segments joining $(0,0)$, $(20,12)$, $(30,12)$, and $(45,6)$. 2. The filling rate is $12\div20=0.6$ gallon per minute. 3. While filling, $0.6t=9$, so $t=15$ minutes. 4. While emptying, volume falls $6$ gallons in $15$ minutes, or $0.4$ gallon per minute. It reaches $9$ gallons $3\div0.4=7.5$ minutes after minute $30$, at $t=37.5$ minutes.

Answer

a) $(0,0)$, $(20,12)$, $(30,12)$, $(45,6)$, joined by straight segments b) $0.6$ gallon per minute c) At $15$ minutes and $37.5$ minutes

Interpret rate of change and initial value 56

A hiking group starts at an elevation of $1500\,\text{ft}$. Its elevation rises at a steady rate to $3500\,\text{ft}$ in $2$ hours. The group rests for $30$ minutes, then descends at a steady rate to a lodge at $2500\,\text{ft}$ over $90$ minutes. a) Describe the elevation-time graph by listing its vertices in hours and feet. b) During which moving phase does elevation change faster per hour? Show the calculations. c) At what times after the start is the group at $3000\,\text{ft}$?

Hints

- Convert the rest and descent times into hours for one time axis. - Compare how many feet the elevation changes in each hour of climbing and descending. - The target elevation might be reached on both the upward and downward segments.

Solution

1. The graph joins $(0,1500)$, $(2,3500)$, $(2.5,3500)$, and $(4,2500)$ by straight line segments. 2. Elevation rises by $\frac{3500-1500}{2}=1000\,\text{ft/h}$ during the climb and falls by $\frac{3500-2500}{1.5}=\frac{2000}{3}\approx666.7\,\text{ft/h}$ during descent. The climb has the larger change per hour. 3. On the climb, $1500+1000t=3000$, so $t=1.5$ hours. 4. On the descent, $3500-\frac{2000}{3}(t-2.5)=3000$, giving $t=3.25$ hours.

Answer

a) $(0,1500)$, $(2,3500)$, $(2.5,3500)$, $(4,2500)$, joined by straight segments b) The climb: $1000\,\text{ft/h}$, compared with approximately $666.7\,\text{ft/h}$ during descent c) After $1.5$ hours and after $3.25$ hours

Interpret rate of change and initial value 57

The table shows the height of a grain plant over $30$ days. <table><tr><td>Time $t$ (days)</td><td>$0$</td><td>$5$</td><td>$15$</td><td>$20$</td><td>$30$</td></tr><tr><td>Height $h$ (cm)</td><td>$12$</td><td>$22$</td><td>$62$</td><td>$92$</td><td>$112$</td></tr></table> a) Find the average growth rate, in centimeters per day, on $[0, 5]$, $[5, 15]$, $[15, 20]$, and $[20, 30]$. b) During which interval did the plant grow fastest? c) Suppose the plant had continued after day $20$ at the rate from $[15, 20]$. What height would it have reached on day $30$? Compare this prediction with the actual value.

Hints

- Divide each height change by the corresponding time change. - The greatest average rate identifies the fastest interval. - For the prediction, start with the day-20 height and add $10$ days of growth.

Solution

1. The rates are: $[0, 5]$: $\frac{22-12}{5}=2\,\text{cm/day}$; $[5, 15]$: $\frac{62-22}{10}=4\,\text{cm/day}$; $[15, 20]$: $\frac{92-62}{5}=6\,\text{cm/day}$; $[20, 30]$: $\frac{112-92}{10}=2\,\text{cm/day}$. 2. The greatest rate is $6\,\text{cm/day}$ on $[15, 20]$. 3. Continuing at $6\,\text{cm/day}$ for $10$ more days gives $92+10(6)=152\,\text{cm}$. The actual height is $112\,\text{cm}$, so growth slowed after day $20$.

Answer

a) $2$, $4$, $6$, and $2\,\text{cm/day}$ b) $[15, 20]$ c) Predicted: $152\,\text{cm}$; actual: $112\,\text{cm}$

Interpret rate of change and initial value 58

A simplified hypothetical data set shows the water depth above the bottom of a reservoir during a drought. <table><tr><td>Day $t$</td><td>$0$</td><td>$5$</td><td>$10$</td><td>$15$</td><td>$20$</td></tr><tr><td>Water depth $h$ (ft)</td><td>$160$</td><td>$155$</td><td>$148$</td><td>$139$</td><td>$128$</td></tr></table> a) List the ordered pairs represented by the data and describe their trend. b) A technician suspects the depth decreases linearly by $1.5\,\text{ft}$ per day. Determine whether this model matches the data exactly. c) Find the average rate of change from day $15$ to day $20$.

Hints

- Compare the changes over equal five-day intervals. - Test the proposed rate against at least one table value. - Average rate of change compares the change in depth with the corresponding change in time.

Solution

1. The ordered pairs are $(0, 160)$, $(5, 155)$, $(10, 148)$, $(15, 139)$, and $(20, 128)$. The depth decreases as time increases, and the decreases become larger over successive five-day intervals. 2. A decrease of $1.5\,\text{ft}$ per day predicts a decrease of $5\cdot1.5=7.5$ feet by day $5$, giving $160-7.5=152.5$ feet. The table gives $155$ feet, so the model does not match exactly. 3. From day $15$ to day $20$, the average rate is $\frac{128-139}{20-15}=\frac{-11}{5}=-2.2\,\text{ft/day}$.

Answer

a) $(0, 160)$, $(5, 155)$, $(10, 148)$, $(15, 139)$, and $(20, 128)$; the water depth decreases at a changing rate. b) No. The model predicts $152.5\,\text{ft}$ on day $5$, but the table gives $155\,\text{ft}$. c) $-2.2\,\text{ft/day}$

Interpret rate of change and initial value 59

Two lines are given: $g_1: y = 1.5x - 3$ $g_2: 2x + y = 4$ a) Find the y-intercept of each line. b) For $g_1$, find the x-value that gives $y = 6$. c) Find the x-intercept of $g_2$. d) Compare the slopes. Which line is increasing and which is decreasing? Briefly explain.

Hints

- What x-value gives a y-intercept? - Substitute the given y-value in part b) and solve. - What y-value gives an x-intercept? - How does the sign of the slope affect a line?

Solution

1. For $g_1$, setting $x = 0$ gives the y-intercept $(0, -3)$. Rewrite $g_2$ as $y = -2x + 4$, so its y-intercept is $(0, 4)$. 2. For $g_1$, solve $6 = 1.5x - 3$. Then $9 = 1.5x$, so $x = 6$. 3. For the x-intercept of $g_2$, set $y = 0$: $2x = 4$, so $x = 2$. Therefore, the x-intercept is $(2, 0)$. 4. The slope of $g_1$ is $1.5 > 0$, so it is increasing. The slope of $g_2$ is $-2 < 0$, so it is decreasing.

Answer

a) $g_1: (0, -3)$; $g_2: (0, 4)$ b) $x = 6$ c) $(2, 0)$ d) $g_1$ is increasing; $g_2$ is decreasing.

Interpret rate of change and initial value 60

Consider the linear function $y = -\frac{2}{3}x + 2$. a) Point $A(6, y_A)$ lies on the graph. Find $y_A$. b) Point $B(x_B, 0)$ also lies on the graph. Find $x_B$. c) Determine algebraically whether $C(-3, 4)$ lies on the line. d) Describe how $y$ changes when $x$ increases by 3 units. Use the slope to justify your answer.

Hints

- If a point lies on a line, what must be true when its coordinates are substituted into the equation? - What does a negative slope tell you about how the values change? - How can you interpret the slope as a fraction relating vertical change to horizontal change? - How can you check whether a point lies on the line without drawing the graph?

Solution

1. For $A$, $y_A = -\frac{2}{3} \cdot 6 + 2 = -2$. 2. For $B$, solve $0 = -\frac{2}{3}x_B + 2$. This gives $x_B = 3$. 3. At $x = -3$, $y = -\frac{2}{3} \cdot (-3) + 2 = 4$, so $C$ lies on the line. 4. With slope $-\frac{2}{3}$, a horizontal increase of $3$ gives $\Delta y = -\frac{2}{3} \cdot 3 = -2$. Thus, $y$ decreases by 2 units.

Answer

a) $y_A = -2$ b) $x_B = 3$ c) Yes. d) $y$ decreases by 2 units.

Interpret rate of change and initial value 61

The water level at a coastal harbor is measured over $24$ hours. <table> <tr><td>Time</td><td>$12{:}00$ a.m.</td><td>$3{:}00$ a.m.</td><td>$6{:}00$ a.m.</td><td>$9{:}00$ a.m.</td><td>$12{:}00$ p.m.</td><td>$3{:}00$ p.m.</td><td>$6{:}00$ p.m.</td><td>$9{:}00$ p.m.</td><td>$12{:}00$ a.m.</td></tr> <tr><td>Water depth (ft)</td><td>$18$</td><td>$24$</td><td>$18$</td><td>$12$</td><td>$18$</td><td>$24$</td><td>$18$</td><td>$12$</td><td>$18$</td></tr> </table> Assume the water depth changes linearly between measurements. 1. Find the water depth at $4{:}30$ a.m. and $10{:}30$ a.m. 2. During which time intervals is the water level rising, and during which is it falling? 3. A fishing boat needs at least $16\,\text{ft}$ of water. Find the time intervals when the boat cannot safely enter or leave the harbor.

Hints

- Use linear interpolation between consecutive measurements. - Compare consecutive depth values to identify rising and falling intervals. - Find where the linear segments cross the minimum safe depth.

Solution

1. At $4{:}30$ a.m., the time is halfway between $3{:}00$ a.m. and $6{:}00$ a.m., so the depth is halfway between $24$ and $18$: $21\,\text{ft}$. At $10{:}30$ a.m., it is halfway between $12$ and $18$: $15\,\text{ft}$. 2. The level rises from $12{:}00$ a.m. to $3{:}00$ a.m., from $9{:}00$ a.m. to $3{:}00$ p.m., and from $9{:}00$ p.m. to $12{:}00$ a.m. It falls from $3{:}00$ a.m. to $9{:}00$ a.m. and from $3{:}00$ p.m. to $9{:}00$ p.m. 3. From $6{:}00$ a.m. to $9{:}00$ a.m., the depth falls from $18$ to $12$, a rate of $-2\,\text{ft/h}$. It reaches $16\,\text{ft}$ at $7{:}00$ a.m. On the rising segment after $9{:}00$ a.m., it reaches $16\,\text{ft}$ at $11{:}00$ a.m. The same pattern repeats $12$ hours later. Because exactly $16\,\text{ft}$ is safe, the unsafe intervals exclude their endpoints.

Answer

1. $4{:}30$ a.m.: $21\,\text{ft}$ $10{:}30$ a.m.: $15\,\text{ft}$ 2. Rising: $12{:}00$ a.m.–$3{:}00$ a.m., $9{:}00$ a.m.–$3{:}00$ p.m., and $9{:}00$ p.m.–$12{:}00$ a.m. Falling: $3{:}00$ a.m.–$9{:}00$ a.m. and $3{:}00$ p.m.–$9{:}00$ p.m. 3. Between $7{:}00$ a.m. and $11{:}00$ a.m., excluding the endpoints, and between $7{:}00$ p.m. and $11{:}00$ p.m., excluding the endpoints.

Interpret rate of change and initial value 62

An electric car's battery level is recorded during a trip. <table> <tr><td>Time</td><td>$8{:}00$ a.m.</td><td>$10{:}00$ a.m.</td><td>$10{:}30$ a.m.</td><td>$12{:}30$ p.m.</td><td>$1{:}30$ p.m.</td></tr> <tr><td>Battery level (%)</td><td>$100$</td><td>$60$</td><td>$90$</td><td>$20$</td><td>$20$</td></tr> </table> 1. Find the average battery use, in percentage points per hour, from $8{:}00$ a.m. to $10{:}00$ a.m. 2. During which interval was the battery-use rate greatest? Justify by comparing rates of change. 3. Give a plausible explanation for the constant battery level from $12{:}30$ p.m. to $1{:}30$ p.m. 4. Suppose that after $1{:}30$ p.m. the battery continues decreasing at the same average rate as from $10{:}30$ a.m. to $12{:}30$ p.m. At what time does the linear model predict a battery level of $0\%$?

Hints

- Find each change in battery level per hour. - Interpret positive, negative, and zero rates in context. - Divide the remaining battery level by the projected use rate.

Solution

1. The level decreases by $100-60=40$ percentage points in $2$ hours, or $20$ percentage points per hour. 2. From $10{:}30$ a.m. to $12{:}30$ p.m., the level decreases by $90-20=70$ percentage points in $2$ hours, or $35$ percentage points per hour. This is greater than the $20$-point rate from $8{:}00$ a.m. to $10{:}00$ a.m. From $10{:}00$ a.m. to $10{:}30$ a.m., the battery is charging rather than being used. 3. The car may have been parked and turned off, so its battery level did not change noticeably. 4. At $35$ percentage points per hour, the remaining $20\%$ lasts $\frac{20}{35}\approx0.5714$ hour, or about $34.3$ minutes. The model predicts the battery will be empty at about $2{:}04$ p.m.

Answer

1. $20$ percentage points per hour 2. $10{:}30$ a.m. to $12{:}30$ p.m., at $35$ percentage points per hour 3. The car may have been parked and turned off. 4. About $2{:}04$ p.m.

Interpret rate of change and initial value 63

A rainwater tank is used to water a garden. The same amount of water is used each day. After $5$ days, $300\,\text{gal}$ remain in the tank. After $12$ days, $230\,\text{gal}$ remain. a) How many gallons of water are used each day? b) How much water was in the tank at the beginning? c) After how many days will the tank be empty?

Hints

- How much water was used between the two measurements? - Divide that change by the number of days between the measurements. - Use the daily rate to work backward from day $5$ to day $0$. - How many daily amounts fit into the starting amount?

Solution

1. Between day $5$ and day $12$, the amount of water decreases by $300 - 230 = 70$ gallons over $12 - 5 = 7$ days. 2. The daily decrease is $70 \div 7 = 10$ gallons per day. 3. Five days before there were $300$ gallons remaining, the tank contained $300 + 5 \cdot 10 = 350$ gallons. 4. At $10$ gallons per day, $350 \div 10 = 35$, so the tank will be empty after $35$ days.

Answer

a) $10\,\text{gal/day}$ b) $350\,\text{gal}$ c) $35$ days

Interpret rate of change and initial value 64

A smartphone battery drains at a constant rate while the phone is in use. At $10{:}00$ a.m., the battery level is $80\%$. At $11{:}20$ a.m., it is $60\%$. a) Find the battery loss in percentage points per hour. b) Assuming the phone has been used continuously at this rate, at what time was it unplugged at $100\%$? c) If the same rate continues, at what time will the battery reach $0\%$?

Hints

- How many minutes pass between the two battery readings? - How many percentage points does the battery lose during that interval? - Find the loss per minute first, then convert it to a per-hour rate. - How long would a $20$-percentage-point drop take at that rate? - How long would a $100$-percentage-point drop take at the same rate?

Solution

1. From $10{:}00$ a.m. to $11{:}20$ a.m. is $80$ minutes. The battery level decreases by $20$ percentage points, so the rate is $20 \div 80 = 0.25$ percentage point per minute, or $0.25 \cdot 60 = 15$ percentage points per hour. 2. Going from $100\%$ to $80\%$ is a decrease of $20$ percentage points, which takes $20 \div 0.25 = 80$ minutes. Eighty minutes before $10{:}00$ a.m. is $8{:}40$ a.m. 3. Going from $100\%$ to $0\%$ takes $100 \div 0.25 = 400$ minutes, or $6$ hours $40$ minutes. Starting at $8{:}40$ a.m., the battery reaches $0\%$ at $3{:}20$ p.m.

Answer

a) $15$ percentage points per hour b) $8{:}40$ a.m. c) $3{:}20$ p.m.

Interpret rate of change and initial value 65

A car travels at a constant speed on the highway. After $20$ minutes, its odometer reads $70\,\text{mi}$. After $50$ minutes total, it reads $100\,\text{mi}$. Find the car's speed in miles per hour and the odometer reading at the start of the measurement period, when $t = 0$.

Hints

- What constant rate connects the two odometer readings? - Find the change in distance divided by the change in time. - How do you convert miles per minute to miles per hour? - Once you know the rate, work backward from the first reading to $t = 0$.

Solution

1. Over the $50 - 20 = 30$ minutes between readings, the odometer increases by $100 - 70 = 30$ miles. The rate is $30 \div 30 = 1$ mile per minute. 2. Convert the rate to miles per hour: $1 \cdot 60 = 60\,\text{mph}$. 3. In the first $20$ minutes, the car travels $20$ miles. Therefore, the odometer reading at $t = 0$ was $70 - 20 = 50\,\text{mi}$.

Answer

The speed is $60\,\text{mph}$, and the odometer reading at $t = 0$ was $50\,\text{mi}$.

Interpret rate of change and initial value 66

Two candles are lit at the same time and each burns at a constant rate. Candle A is $8\,\text{in.}$ tall after $2$ hours and $5\,\text{in.}$ tall after $5$ hours. Candle B is $9\,\text{in.}$ tall after $3$ hours and $5.4\,\text{in.}$ tall after $6$ hours. Use calculations to determine which candle was taller at $t = 0$ and which candle burns faster, in inches per hour.

Hints

- What does a constant burn rate tell you about the type of relationship? - For each candle, use the change in height divided by the change in time to find the burn rate. - Once you know each rate, work backward to find each starting height. - Compare the two starting heights and the two burn rates.

Solution

1. For Candle A, the height decreases by $8 - 5 = 3$ inches over $5 - 2 = 3$ hours, so it burns at $1\,\text{in./h}$. Working back $2$ hours gives an initial height of $8 + 2 \cdot 1 = 10$ inches. 2. For Candle B, the height decreases by $9 - 5.4 = 3.6$ inches over $6 - 3 = 3$ hours, so it burns at $1.2\,\text{in./h}$. Working back $3$ hours gives an initial height of $9 + 3 \cdot 1.2 = 12.6$ inches. 3. Candle B started taller because $12.6 > 10$, and it burns faster because $1.2 > 1$.

Answer

Candle B was taller at the start: $12.6\,\text{in.}$ compared with Candle A's $10\,\text{in.}$. Candle B also burns faster: $1.2\,\text{in./h}$ compared with $1\,\text{in./h}$.

Interpret rate of change and initial value 67

A full water tank is drained at a constant rate. The water height $h$, in centimeters, after $t$ minutes is modeled by $h=120-8t$. a) Find the water height at $t=0$, $t=5$, $t=10$, and $t=15$. Record the results in a table. b) What does $120$ represent in this situation? c) After how many minutes is the tank empty? Justify your answer using the equation.

Hints

- Substitute each given time into the equation. - What time value represents the moment draining begins? - What water height represents an empty tank?

Solution

1. Substitute each time into $h=120-8t$: $t=0$: $h=120-8\cdot0=120$ $t=5$: $h=120-8\cdot5=80$ $t=10$: $h=120-8\cdot10=40$ $t=15$: $h=120-8\cdot15=0$ 2. The constant $120$ is the initial water height, because it is the value of $h$ when $t=0$. 3. The tank is empty when $h=0$. Since $120-8\cdot15=0$, the tank is empty after $15$ minutes.

Answer

a) <table><thead><tr><th>Time $t$ (min)</th><th>Height $h$ (cm)</th></tr></thead><tbody><tr><td>$0$</td><td>$120$</td></tr><tr><td>$5$</td><td>$80$</td></tr><tr><td>$10$</td><td>$40$</td></tr><tr><td>$15$</td><td>$0$</td></tr></tbody></table> b) $120$ is the initial water height in centimeters. c) $15$ minutes, because substituting $t=15$ gives $h=0$.

Interpret rate of change and initial value 68

a) For panels a), b), and c), find the temperature at noon and the change in temperature per hour. Include units. b) Which model shows warming, which shows cooling, and which shows no change? c) Why can't panel d) represent temperature as a function of time?

Hints

- What horizontal-axis value means noon? - Compare temperatures at two times one hour apart to find the hourly change. - Check whether panel d) gives one temperature or several for the same time.

Solution

1. At noon, the input is $0$. Panel a) shows $-1\,^\circ\text{F}$, and the temperature rises $2\,^\circ\text{F}$ per hour. 2. Panel b) starts at $2\,^\circ\text{F}$ at noon and falls $0.5\,^\circ\text{F}$ per hour, a rate of $-0.5\,^\circ\text{F}$ per hour. 3. Panel c) stays at $-3\,^\circ\text{F}$, so its rate is $0\,^\circ\text{F}$ per hour. 4. Model a) shows warming, b) cooling, and c) no change. 5. Panel d) shows several temperatures at the same time, two hours before noon, so it cannot be a function of time.

Answer

a) Panel a): $-1\,^\circ\text{F}$ at noon, $+2\,^\circ\text{F}$ per hour; panel b): $2\,^\circ\text{F}$, $-0.5\,^\circ\text{F}$ per hour; panel c): $-3\,^\circ\text{F}$, $0\,^\circ\text{F}$ per hour. b) a) warming; b) cooling; c) no change. c) One time has more than one temperature in panel d).

Interpret rate of change and initial value 69

A large water tank is being drained at a constant rate. The graph shows the remaining volume $V$, in gallons, after $t$ minutes. a) Choose two convenient points from the graph and use them to find the slope $m$. Explain what the slope means in this situation. b) Find the y-intercept $b$ from the graph and write the linear function in the form $V(t) = mt + b$. c) After how many minutes will the tank be empty?

Hints

- Look for points on the line that lie exactly on grid intersections. - Use change in volume divided by change in time to find the slope. - What volume does the graph show when $t = 0$? - How do the slope and initial value fit into $V(t) = mt + b$? - What value of $V$ represents an empty tank?

Solution

1. Two convenient points are $(0, 250)$ and $(5, 150)$. The slope is $m = \frac{150 - 250}{5 - 0} = -20\,\text{gal/min}$. This means the amount of water decreases by $20$ gallons each minute. 2. The graph crosses the y-axis at $(0, 250)$, so $b = 250$. Therefore, $V(t) = -20t + 250$. 3. The tank is empty when $V(t) = 0$. Solve $0 = -20t + 250$, giving $t = 12.5$ minutes.

Answer

a) $m = -20\,\text{gal/min}$; the tank loses $20$ gallons of water per minute. b) $V(t) = -20t + 250$ c) $12.5$ minutes

Interpret rate of change and initial value 70

During a $12$-minute chemistry experiment, the temperature of a liquid is recorded while it is heated and cooled. a) Find the initial temperature and the maximum temperature. b) How long did the temperature remain constant at its maximum? c) Find the average rate of temperature change during the first $3$ minutes, in degrees Celsius per minute. d) At what times was the temperature exactly $65\,^{\circ}\text{C}$?

Hints

- Read the starting point and highest point of the graph. - A horizontal segment represents a constant temperature. - Average rate of change is change in temperature divided by change in time. - For d), write a linear equation for each segment that crosses $65\,^{\circ}\text{C}$.

Solution

1. At $t = 0$, the temperature is $20\,^{\circ}\text{C}$. The maximum is $80\,^{\circ}\text{C}$. 2. The graph is horizontal at $80\,^{\circ}\text{C}$ from $t = 3$ to $t = 6$, so the maximum lasts $6 - 3 = 3$ minutes. 3. During the first $3$ minutes, the average rate of change is $\frac{80 - 20}{3 - 0} = 20\,^{\circ}\text{C}/\text{min}$. 4. On the heating segment, $T = 20 + 20t$. Solving $20 + 20t = 65$ gives $t = 2.25$ minutes. 5. On the first cooling segment, $T = 80 - 15(t - 6)$. Solving $80 - 15(t - 6) = 65$ gives $t = 7$ minutes.

Answer

a) Initial temperature: $20\,^{\circ}\text{C}$; maximum temperature: $80\,^{\circ}\text{C}$ b) $3$ minutes c) $20\,^{\circ}\text{C}/\text{min}$ d) $t = 2.25$ minutes and $t = 7$ minutes

Interpret rate of change and initial value 71

A hot-air balloon takes a multi-hour flight. The graph shows altitude as a function of time since takeoff. a) Give the time intervals during which the balloon remains at a constant altitude and state each altitude. b) At what times is the balloon exactly $1200\,\text{ft}$ above the ground? c) Find the ascent rate during the first hour. d) During which interval does the balloon descend fastest? Find the descent rate.

Hints

- Horizontal graph segments represent constant altitude. - Check both an ascending segment and a descending segment for the requested altitude. - Compute each rate as change in altitude divided by change in time. - The steepest downward segment has the greatest descent-rate magnitude.

Solution

1. The balloon remains at $900\,\text{ft}$ for $t\in[1, 1.5]$ hours and at $1800\,\text{ft}$ for $t\in[3, 4]$ hours. 2. From $t=1.5$ to $t=3$, the altitude rises from $900$ to $1800$ feet at $600\,\text{ft/h}$, so it reaches $1200\,\text{ft}$ at $t=2$ hours. From $t=4$ to $t=5$, the altitude falls from $1800$ to $600$ feet at $-1200\,\text{ft/h}$, so it reaches $1200\,\text{ft}$ at $t=4.5$ hours. 3. During the first hour, the balloon rises $900$ feet, so the ascent rate is $900\,\text{ft/h}$. 4. The steepest descending segment is from $t=4$ to $t=5$, where the balloon drops $1200$ feet in $1$ hour. Its descent rate is $1200\,\text{ft/h}$ downward.

Answer

a) $t\in[1, 1.5]$ h at $900\,\text{ft}$; $t\in[3, 4]$ h at $1800\,\text{ft}$ b) $t=2$ hours and $t=4.5$ hours c) $900\,\text{ft/h}$ d) From $t=4$ to $t=5$ hours; $1200\,\text{ft/h}$ downward

Interpret rate of change and initial value 72

The graph shows the temperature in a mountain town from 6:00 a.m. to 4:00 p.m. A cold front moved in late that morning. Time $t=0$ means 6:00 a.m. a) What were the clock time and temperature at $t=2$ and $t=8$? b) At what values of $t$ was the temperature $0\,^{\circ}\text{F}$? Give the zeros and their x-intercept points. c) How many degrees did the temperature drop from its highest value to the end of the graph? d) Which interval had the greater average warming rate: $t=0$ to $t=2$, or $t=2$ to $t=5$? Find both rates.

Hints

- The horizontal axis measures hours since 6:00 a.m., not clock hours. - Look for places where the temperature changes from negative to positive or positive to negative. - Include the part of the drop below zero when comparing the highest and final temperatures. - For each interval, divide the change in temperature by the number of hours.

Solution

1. At $t=2$, it was 8:00 a.m. and $4\,^{\circ}\text{F}$. At $t=8$, it was 2:00 p.m. and $-2\,^{\circ}\text{F}$. 2. The graph crosses the x-axis at $t=1$ and $t=7.5$. The zeros are $1$ and $7.5$; the x-intercept points are $(1, 0)$ and $(7.5, 0)$. 3. The highest temperature is $10\,^{\circ}\text{F}$. At $t=10$, it is $-6\,^{\circ}\text{F}$. The drop is $10-(-6)=16\,^{\circ}\text{F}$. 4. From $t=0$ to $t=2$, the average rate is $\frac{4-(-4)}{2}=4\,^{\circ}\text{F/h}$. From $t=2$ to $t=5$, it is $\frac{10-4}{3}=2\,^{\circ}\text{F/h}$. The first interval has the greater average warming rate.

Answer

a) $t=2$: 8:00 a.m., $4\,^{\circ}\text{F}$; $t=8$: 2:00 p.m., $-2\,^{\circ}\text{F}$ b) Zeros: $1$ and $7.5$; x-intercepts: $(1, 0)$ and $(7.5, 0)$ c) $16\,^{\circ}\text{F}$ d) $t=0$ to $t=2$: $4\,^{\circ}\text{F/h}$; $t=2$ to $t=5$: $2\,^{\circ}\text{F/h}$. The first interval.

Interpret rate of change and initial value 73

A water tank is filled at a constant rate. The function $V(t)$ gives the volume in gallons after $t$ minutes. Two measurements, $P_1$ and $P_2$, are shown on the graph. a) Find the inflow rate, or slope, in gallons per minute. b) Find the volume already in the tank at $t=0$. c) Write the function $V(t)$. d) Find the volume after $15$ minutes.

Hints

- Use change in volume divided by change in time to find the slope. - Substitute one point into $V(t)=mt+b$ to find the initial value. - Evaluate the completed function at $t=15$.

Solution

1. The points are $P_1(2, 100)$ and $P_2(8, 250)$. The slope is $m=\frac{250-100}{8-2}=\frac{150}{6}=25\,\text{gal/min}$. 2. Substitute $P_1$ into $V(t)=25t+b$: $100=25\cdot2+b$, so $b=50$ gallons. 3. The function is $V(t)=25t+50$. 4. $V(15)=25\cdot15+50=425$, so the tank contains $425$ gallons.

Answer

a) $25\,\text{gal/min}$ b) $50$ gallons c) $V(t)=25t+50$ d) $425$ gallons

Interpret rate of change and initial value 74

Consider the graph of the linear function $f$. a) Find its slope $m$ and y-intercept $b$. b) Find its zero and state the corresponding x-intercept point. c) Write a proportional function $p$ with the same slope.

Hints

- Read the output when the input is $0$ for the y-intercept. - Use two grid points to find rise divided by run. - A zero is an input; the x-intercept is the point with output $0$. - A proportional function's graph passes through the origin.

Solution

1. The graph crosses the y-axis at $(0,-3)$. Using the point $(4,0)$, its slope is $m=\frac{0-(-3)}{4-0}=\frac{3}{4}=0.75$; thus $b=-3$. 2. The function is $f(x)=0.75x-3$. Solving $f(x)=0$ gives $x=4$, so the x-intercept is $(4,0)$. 3. A proportional function starts at $0$. With the same rate, $p(x)=0.75x$.

Answer

a) $m=0.75$; $b=-3$. b) Zero: $x=4$; x-intercept point: $(4,0)$. c) $p(x)=0.75x$

Interpret rate of change and initial value 75

Two cell phone plans have different monthly fees and different per-minute charges. Let $x$ be the number of minutes used in a month. On the graph, curve $a$ represents Plan A and curve $b$ represents Plan B. Write a cost function for each plan, including the contextual domain. What does the intersection of the two lines mean in context?

Hints

- The y-intercept represents the monthly fee. - The slope represents the cost per minute. - The number of minutes used cannot be negative. - At an intersection, both plans have the same cost for the same number of minutes.

Solution

1. Plan A has a y-intercept of $5$, representing a $\$5$ monthly fee. Using the point $(100, 15)$, its slope is $m=\frac{15-5}{100-0}=0.10$. Thus, $a(x)=0.10x+5$, with $x\ge0$. 2. Plan B has a y-intercept of $10$, representing a $\$10$ monthly fee. Using $(100, 15)$, its slope is $m=\frac{15-10}{100-0}=0.05$. Thus, $b(x)=0.05x+10$, with $x\ge0$. 3. The lines intersect at $(100, 15)$. This means both plans cost $\$15$ when $100$ minutes are used.

Answer

Plan A: $a(x)=0.10x+5$, $x\ge0$; Plan B: $b(x)=0.05x+10$, $x\ge0$; at $100$ minutes, both plans cost $\$15$.

Interpret rate of change and initial value 76

A measuring container is made from two cylindrical sections stacked directly on top of each other. The lower section is $8\,\text{cm}$ tall. Water flows into the container at a constant volume rate. The graph shows water height $h$, in centimeters, as a function of time $t$, in seconds. a) After how many seconds is the lower cylindrical section full? b) In which section does the water height rise faster? What does this tell you about the diameters of the two cylinders? c) Find the slope on each graph segment. What does each slope mean in context?

Hints

- The change in slope occurs when the water reaches the upper cylinder. - Compute slope as change in height divided by change in time. - With the same inflow, water rises faster in a narrower cylinder.

Solution

1. The lower section is full when the water reaches $8\,\text{cm}$. The graph reaches this height at $t=4\,\text{s}$. 2. In the lower section, the water rises $8\,\text{cm}$ in $4\,\text{s}$, or $2\,\text{cm/s}$. In the upper section, it rises from $8$ to $18\,\text{cm}$ in $2\,\text{s}$, or $5\,\text{cm/s}$. The water rises faster in the upper section, so that cylinder has a smaller cross-sectional area and therefore a smaller diameter. 3. The first slope is $m_1=\frac{8-0}{4-0}=2\,\text{cm/s}$. The second slope is $m_2=\frac{18-8}{6-4}=5\,\text{cm/s}$. Each slope is the rate at which the water height increases.

Answer

a) $4\,\text{s}$ b) The upper section; it has the smaller diameter. c) Lower section: $2\,\text{cm/s}$; upper section: $5\,\text{cm/s}$

Interpret rate of change and initial value 77

Lara rides her bicycle to a lake. The graph shows her distance from home. a) How far is the lake from her home? b) After how many minutes does she stop for a break, and how long does the break last? c) On which riding segment or segments does she travel fastest? Find her speed in miles per hour.

Hints

- The lake is represented by the greatest distance from home. - A horizontal segment represents a stop. - Convert minutes to hours before calculating miles per hour.

Solution

1. The greatest distance shown is $9$ miles, so the lake is $9$ miles from her home. 2. The horizontal segment begins after $30$ minutes and ends after $45$ minutes. The break lasts $15$ minutes. 3. On the first segment, Lara travels $6$ miles in $30$ minutes, or $0.5$ hour, so her speed is $\frac{6}{0.5}=12\,\text{mph}$. On the last segment, she travels $3$ miles in $15$ minutes, or $0.25$ hour, so her speed is $\frac{3}{0.25}=12\,\text{mph}$. She travels equally fast on both riding segments.

Answer

a) $9$ miles b) After $30$ minutes; the break lasts $15$ minutes. c) Both riding segments; $12\,\text{mph}$

Interpret rate of change and initial value 78

The graph shows the elevation profile of an $18$-mile bicycle route through hilly terrain. a) Find the starting elevation and the maximum elevation reached. b) Give the part of the route, from mile ___ to mile ___, where the riders travel downhill. c) A rider claims, “The climb from mile $3$ to mile $6$ has twice the average elevation gain per mile of the climb from mile $12$ to mile $18$.” Check the claim by comparing the average elevation gain per mile on the two intervals.

Hints

- Read the first and highest points on the graph. - A decreasing graph segment represents downhill travel. - Divide elevation gain by horizontal distance to compare average steepness.

Solution

1. The route starts at $500\,\text{ft}$ and reaches a maximum elevation of $1500\,\text{ft}$ at the end. 2. The graph decreases from mile $6$ to mile $12$, so that is the downhill section. 3. From mile $3$ to mile $6$, the elevation gain is $1200-600=600\,\text{ft}$ over $3$ miles, or $200\,\text{ft/mi}$. From mile $12$ to mile $18$, the gain is $1500-900=600\,\text{ft}$ over $6$ miles, or $100\,\text{ft/mi}$. The middle climb is twice as steep on average, so the claim is correct.

Answer

a) Starting elevation: $500\,\text{ft}$; maximum elevation: $1500\,\text{ft}$ b) Mile $6$ to mile $12$ c) The claim is correct: $200\,\text{ft/mi}$ versus $100\,\text{ft/mi}$.

Interpret rate of change and initial value 79

A candle is $8\,\text{in.}$ tall when it is lit. Its height $h$, in inches, decreases at a constant rate as it burns for $t$ hours. a) Use the graph to write the linear function $h(t)$. Pay attention to the scale on each axis. b) Find the candle's height after $12$ hours and check the value on the graph.

Hints

- Since the candle gets shorter, what sign should the slope have? - Check how much one grid step represents on each axis. - Use two easy-to-read points to find the change in height per hour. - Substitute the requested time into your function.

Solution

1. Two convenient points are $(0, 8)$ and $(8, 4)$. 2. The slope is $m = \frac{4 - 8}{8 - 0} = -\frac{1}{2}$. The initial height is $8$, so $h(t) = -\frac{1}{2}t + 8$. 3. $h(12) = -\frac{1}{2} \cdot 12 + 8 = 2$. The graph also shows $(12, 2)$.

Answer

a) $h(t) = -\frac{1}{2}t + 8$ b) $2\,\text{in.}$

Interpret rate of change and initial value 80

A rainwater tank fills at a constant rate during a 15-minute observation. The graph shows the amount of water in the tank. a) How much water is in the tank at the start? b) How many gallons are added each minute? Explain how you found the rate from the graph. c) Write a rule for the volume $V(t)$ after $t$ minutes, and find when the tank reaches $90$ gallons.

Hints

- Read the volume where time is zero. - Choose two points on the line and compare the change in volume with the change in time. - A constant-rate rule combines the starting amount with the amount added over time.

Solution

1. The graph starts at $(0,25)$, so the initial volume is $25$ gallons. 2. At $t=5$, the volume is $50$ gallons. The rate is $(50-25)\div5=5$ gallons per minute. 3. The volume is $V(t)=25+5t$ for $0\le t\le15$. Solving $25+5t=90$ gives $t=13$ minutes.

Answer

a) $25$ gallons b) $5$ gallons per minute, using the rise over a $5$-minute interval. c) $V(t)=25+5t$ for $0\le t\le15$; $13$ minutes

Interpret rate of change and initial value 81

A candle burns down at a constant rate. The graph shows its height over time. a) What was the candle's height when it was lit? b) How much height does it lose per hour? Explain using two points on the graph. c) Write the height rule $h(t)$, and find when its height reaches $4.5$ inches.

Hints

- The height at time zero is the starting height. - Compare how far the height falls with the time elapsed. - In the rule, a loss each hour must reduce the starting amount.

Solution

1. The graph meets the vertical axis at $12$, so the starting height is $12$ inches. 2. The height goes from $12$ inches at $0$ hours to $9$ inches at $4$ hours, a change of $-3$ inches in $4$ hours. Its rate is $-3\div4=-0.75$ inch per hour. 3. The rule is $h(t)=12-0.75t$. Solving $12-0.75t=4.5$ gives $t=10$ hours.

Answer

a) $12$ inches b) It loses $0.75$ inch per hour; the rate is $-0.75$ inch per hour. c) $h(t)=12-0.75t$; $10$ hours

Interpret rate of change and initial value 82

The graph shows two functions. One represents a proportional relationship. Identify that function and write its equation. Also write the equation of the other line.

Hints

- What graph feature identifies a proportional relationship? - A proportional line has a y-intercept of zero. - Use rise over run to find each slope.

Solution

1. A proportional graph passes through $(0, 0)$, so line $p$ is proportional. 2. Line $p$ also passes through $(2, 3)$, so its slope is $3 \div 2=1.5$. Its equation is $y=1.5x$. 3. Line $q$ crosses the y-axis at $3$ and passes through $(3, 0)$. Its slope is $(0-3) \div (3-0)=-1$, so its equation is $y=-x+3$.

Answer

Line $p$ is proportional and has equation $y=1.5x$. Line $q$ has equation $y=-x+3$.

Interpret rate of change and initial value 83

A rainwater tank is observed for $10$ hours. The graph shows the volume of water in the tank. a) How much water is in the tank at the start? b) During which interval does the heaviest rain occur? Justify your answer by comparing slopes. c) After the rain stops, how many gallons are removed for watering?

Hints

- The initial value is the y-intercept. - A steeper upward segment represents a faster increase in volume. - The amount removed is the decrease from the post-rain maximum to the final volume.

Solution

1. At $t=0$, the tank contains $50$ gallons. 2. From $t=0$ to $t=2$, the volume increases by $50$ gallons in $2$ hours, a rate of $25\,\text{gal/h}$. From $t=2$ to $t=3$, it increases by $50$ gallons in $1$ hour, a rate of $50\,\text{gal/h}$. The heaviest rain occurs from hour $2$ to hour $3$. 3. After the rain, the volume eventually decreases from $150$ gallons to $50$ gallons. The amount removed is $150-50=100$ gallons.

Answer

a) $50$ gallons b) From hour $2$ to hour $3$, because the slope is $50\,\text{gal/h}$ rather than $25\,\text{gal/h}$. c) $100$ gallons

Interpret rate of change and initial value 84

A specially shaped glass vase is filled with water at a constant volume rate. The graph shows water height, in centimeters, as a function of time, in seconds. 1. What water height is reached after $2$ seconds? 2. After how many seconds is the water height $9\,\text{cm}$? 3. Over what height range is the vase narrowest? Justify your answer using the graph.

Hints

- With constant inflow, a steeper graph means the water height rises faster. - Water rises fastest where the vase has the smallest cross-sectional area.

Solution

1. The graph passes through $(2, 1)$, so the water is $1\,\text{cm}$ high after $2$ seconds. 2. The graph reaches $9\,\text{cm}$ at $t=6$ seconds. 3. With constant inflow, the water rises fastest where the vase is narrowest. The steepest segment is from $t=2$ to $t=6$, corresponding to water heights from $1\,\text{cm}$ to $9\,\text{cm}$.

Answer

1. $1\,\text{cm}$ 2. $6$ seconds 3. From $1\,\text{cm}$ to $9\,\text{cm}$

Interpret rate of change and initial value 85

A rain-collection basin fills during a short storm and is later drained through a valve. The graph shows water volume $V$ as a function of time $t$. 1) How much water is in the basin after $2$ minutes? 2) How long does the volume remain unchanged before draining begins? 3) Find the fill rate during the storm in gallons per minute. 4) During draining, at what time does the basin contain exactly $5$ gallons?

Hints

- Identify the filling, waiting, and draining phases. - Rate of change is change in volume divided by change in time. - Use the constant drain rate to work backward from $20$ gallons.

Solution

1. The volume increases linearly from $0$ to $20$ gallons in $4$ minutes. After $2$ minutes, the volume is $10$ gallons. 2. The graph is horizontal from $t=4$ to $t=10$, so the volume remains unchanged for $6$ minutes. 3. The fill rate is $\frac{20}{4}=5\,\text{gal/min}$. 4. Draining begins at $t=10$ with $20$ gallons and ends at $t=14$. The drain rate is $5\,\text{gal/min}$. Reaching $5$ gallons requires a decrease of $15$ gallons, which takes $3$ minutes. Therefore, the time is $t=13$ minutes.

Answer

1) $10$ gallons 2) $6$ minutes 3) $5\,\text{gal/min}$ 4) $13$ minutes

Interpret rate of change and initial value 86

The graph shows the elevation profile of a drone surveying a valley. The horizontal axis gives distance traveled in miles, and the vertical axis gives elevation above sea level in feet. 1. At what elevations does the drone start and finish, at $0$ miles and $12$ miles? 2. Over what part of the route does the drone fly at a constant elevation? 3. What maximum elevation does the drone reach? 4. What is the drone's elevation after exactly $10$ miles? 5. Which climb is steeper: from mile $0$ to mile $2$, or from mile $4$ to mile $6$? Justify your answer.

Hints

- Horizontal segments represent constant elevation. - Use the highest point for the maximum. - On a straight segment, calculate the constant elevation change per mile. - Compare rise divided by run for the two climbs.

Solution

1. The drone starts at $1500\,\text{ft}$ and finishes at $1300\,\text{ft}$. 2. The graph is horizontal from mile $2$ to mile $4$, at $2100\,\text{ft}$. 3. The maximum elevation is $2900\,\text{ft}$ at mile $6$. 4. From mile $9$ to mile $12$, elevation decreases from $1900$ to $1300$ feet, a rate of $-200\,\text{ft/mi}$. At mile $10$, the elevation is $1900-200=1700\,\text{ft}$. 5. From mile $0$ to mile $2$, the average climb is $\frac{2100-1500}{2}=300\,\text{ft/mi}$. From mile $4$ to mile $6$, it is $\frac{2900-2100}{2}=400\,\text{ft/mi}$. The second climb is steeper.

Answer

1. Start: $1500\,\text{ft}$; finish: $1300\,\text{ft}$ 2. Mile $2$ to mile $4$ 3. $2900\,\text{ft}$ 4. $1700\,\text{ft}$ 5. Mile $4$ to mile $6$; $400\,\text{ft/mi}$ versus $300\,\text{ft/mi}$

Interpret rate of change and initial value 87

The graph shows a regional train trip between two cities. The x-axis gives time $t$ in minutes since departure, and the y-axis gives distance $s$ from the starting station in miles. 1) How far from the start is the train after exactly one hour? 2) How many total minutes does the train remain stopped at intermediate stations? 3) What is the train's speed, in miles per hour, on the first segment before the first stop? 4) After how many minutes is the train first $60$ miles from the start?

Hints

- Determine which graph segment contains the requested time. - Horizontal segments represent station stops. - Convert minutes to hours before calculating miles per hour. - Use the constant rate on the second moving segment.

Solution

1. At $t=60$, the train is on the segment from $(50, 40)$ to $(90, 80)$, which has a rate of $1\,\text{mi/min}$. Ten minutes after $t=50$, the distance is $40+10=50$ miles. 2. The train is stopped from minute $40$ to minute $50$ and from minute $90$ to minute $100$. The total stop time is $10+10=20$ minutes. 3. On the first segment, the train travels $40$ miles in $40$ minutes, or $\frac{2}{3}$ hour. Its speed is $40\div\frac{2}{3}=60\,\text{mph}$. 4. Starting at minute $50$, the train is $40$ miles from the start and travels at $1\,\text{mi/min}$. It needs $20$ more minutes to reach $60$ miles, so the time is $70$ minutes.

Answer

1) $50$ miles 2) $20$ minutes 3) $60\,\text{mph}$ 4) $70$ minutes

Interpret rate of change and initial value 88

The graph shows the water level $h$, in centimeters, in a rainwater tank over $10$ hours, where $t$ is time in hours. 1) What is the water level after $6$ hours? 2) For how many hours does the water level remain constant? 3) At what rate, in centimeters per hour, does the water level rise during the first $2$ hours? 4) While the tank is filling, at what time is the water level exactly $160\,\text{cm}$?

Hints

- A horizontal graph segment represents no change. - Find a rate of change by dividing the change in water level by the change in time. - Read the initial water level at $t = 0$. - For the final question, identify the segment where the graph passes from below to above $160\,\text{cm}$.

Solution

1. From $t = 4$ to $t = 8$, the graph passes through $(4, 100)$ and $(8, 180)$. Its rate of change is $\frac{180 - 100}{8 - 4} = 20\,\text{cm/h}$. Therefore, $h(6) = 100 + 2 \cdot 20 = 140\,\text{cm}$. 2. The graph is horizontal from $t = 2$ to $t = 4$, so the water level remains constant for $2$ hours. 3. During the first $2$ hours, the level rises from $40\,\text{cm}$ to $100\,\text{cm}$. The rate is $\frac{100 - 40}{2 - 0} = 30\,\text{cm/h}$. 4. The level reaches $160\,\text{cm}$ between $t = 4$ and $t = 8$. Solve $100 + 20(t - 4) = 160$. Then $20(t - 4) = 60$, so $t = 7$.

Answer

1) The water level is $140\,\text{cm}$. 2) It remains constant for $2$ hours. 3) The water level rises at $30\,\text{cm/h}$. 4) The water level is $160\,\text{cm}$ at $t = 7\,\text{h}$.

Interpret rate of change and initial value 89

The graph shows the elevation profile of an $8$-hour mountain hike. The variable $t$ is time in hours, and $h(t)$ is elevation above sea level in feet. 1. What are the group's elevations after $2$ hours and after $6$ hours? 2. During what times does the group remain at its maximum elevation of $4200\,\text{ft}$? 3. How many feet of total elevation gain does the group complete during the first $4$ hours? 4. During which interval does the group descend fastest?

Hints

- Read values from the graph at the requested times. - A horizontal segment means the elevation is unchanged. - For the first four hours, subtract the starting elevation from the ending elevation. - Compare the rates of change on the descending segments.

Solution

1. The graph gives $h(2)=3000\,\text{ft}$. At $t=6$, the point is halfway between $(5, 4200)$ and $(7, 2400)$, so $h(6)=3300\,\text{ft}$. 2. The group remains at the maximum elevation from $t=4$ to $t=5$ hours. 3. The elevation rises continuously from $1200$ feet at the start to $4200$ feet at $t=4$. The total gain is $4200-1200=3000\,\text{ft}$. 4. From $t=5$ to $t=7$, the descent rate is $\frac{2400-4200}{7-5}=-900\,\text{ft/h}$. From $t=7$ to $t=8$, it is $\frac{1200-2400}{8-7}=-1200\,\text{ft/h}$. The group descends fastest from hour $7$ to hour $8$.

Answer

1. After $2$ h: $3000\,\text{ft}$; after $6$ h: $3300\,\text{ft}$ 2. From $4$ to $5$ hours 3. $3000\,\text{ft}$ 4. From hour $7$ to hour $8$

Interpret rate of change and initial value 90

The graph shows a cyclist's speed during a $30$-minute training ride. a) What is the initial value of the speed function, and what does it mean in context? b) Find the rate of change of speed on each interval: $0$–$4$, $4$–$12$, $12$–$20$, $20$–$22$, and $22$–$30$ minutes. Give units. c) On which intervals is the cyclist speeding up, maintaining a constant speed, and slowing down? Explain how the signs of the rates support your answer.

Hints

- The initial value is the graph's output when time is $0$. - For each straight segment, compare the change in speed with the elapsed time. - Positive, zero, and negative rates correspond to three different behaviors of the speed function.

Solution

1. At $t=0$, the graph has speed $0\,\text{mph}$. The initial value is therefore $0\,\text{mph}$, meaning the cyclist starts from rest. 2. From $0$ to $4$ minutes, the rate is $\frac{8-0}{4-0}=2\,\frac{\text{mph}}{\text{min}}$. 3. From $4$ to $12$ minutes, the speed stays at $8\,\text{mph}$, so the rate is $0\,\frac{\text{mph}}{\text{min}}$. 4. From $12$ to $20$ minutes, the rate is $\frac{16-8}{20-12}=1\,\frac{\text{mph}}{\text{min}}$. 5. From $20$ to $22$ minutes, the rate is $0\,\frac{\text{mph}}{\text{min}}$. 6. From $22$ to $30$ minutes, the rate is $\frac{0-16}{30-22}=-2\,\frac{\text{mph}}{\text{min}}$. 7. Positive rates on $0$–$4$ and $12$–$20$ mean the cyclist is speeding up. Zero rates on $4$–$12$ and $20$–$22$ mean constant speed. The negative rate on $22$–$30$ means the cyclist is slowing down.

Answer

a) $0\,\text{mph}$; the cyclist starts from rest. b) $0$–$4$: $2\,\frac{\text{mph}}{\text{min}}$; $4$–$12$: $0$; $12$–$20$: $1\,\frac{\text{mph}}{\text{min}}$; $20$–$22$: $0$; $22$–$30$: $-2\,\frac{\text{mph}}{\text{min}}$ c) Speeding up: $0$–$4$ and $12$–$20$; constant speed: $4$–$12$ and $20$–$22$; slowing down: $22$–$30$.

Interpret rate of change and initial value 91

A car starts with a full $15\,\text{gal}$ fuel tank. The graph shows the amount of fuel $y$, in gallons, after the car has traveled $x$ miles. 1) How much fuel remains after the car has traveled $200$ miles? 2) After how many miles is the tank empty? 3) How many gallons does the car use, on average, for every $100$ miles traveled? 4) Write a function that gives $y$ in terms of $x$ for this trip.

Hints

- Read the graph value directly above the requested distance. - The x-intercept represents the distance when no fuel remains. - Divide the total fuel used by the total distance, then scale the rate to $100$ miles. - Use $y=mx+b$, where $b$ is the starting amount of fuel.

Solution

1. At $x=200$, the graph has value $y=9$, so $9\,\text{gal}$ remain. 2. The graph reaches the x-axis at $x=500$, so the tank is empty after $500$ miles. 3. The car uses $15\,\text{gal}$ over $500$ miles. The rate is $15\div500=0.03\,\text{gal/mi}$. Over $100$ miles, it uses $0.03\cdot100=3\,\text{gal}$. 4. The initial value is $15$, and the slope is $\frac{0-15}{500-0}=-0.03$. Therefore, $y=-0.03x+15$ for $0\le x\le500$.

Answer

1) $9\,\text{gal}$ 2) $500$ miles 3) $3\,\text{gal}$ per $100$ miles 4) $y=-0.03x+15$, for $0\le x\le500$

Interpret rate of change and initial value 92

Two smartphones, A and B, are connected to different chargers at the same time. The graph labels Phone A's charge as $a$ and Phone B's charge as $b$. The y-value is battery charge, as a percent, after $x$ minutes. 1) Which phone had the greater charge at the beginning? Give both initial values. 2) After how many minutes do the phones have the same charge? What is that charge? 3) What is Phone A's charge after $50$ minutes? 4) Write a function for the charge of each phone.

Hints

- Initial values occur when $x=0$. - The phones have the same charge where the two graphs intersect. - Find each rate of change from two points on its line. - Use the form $y=mx+b$, where $b$ is the initial charge.

Solution

1. The y-intercepts are $10\%$ for Phone A and $40\%$ for Phone B, so Phone B starts with the greater charge. 2. The lines intersect at $(30, 55)$. After $30$ minutes, both phones have a charge of $55\%$. 3. Phone A's rate of change is $\frac{100-10}{60-0}=1.5$ percentage points per minute. After $50$ minutes, its charge is $10+1.5\cdot50=85\%$. 4. For Phone A, the initial value is $10$ and the rate is $1.5$, so $a(x)=1.5x+10$ for $0\le x\le60$. For Phone B, the initial value is $40$ and the rate is $\frac{90-40}{100}=0.5$, so $b(x)=0.5x+40$ for $0\le x\le100$.

Answer

1) Phone B; Phone A starts at $10\%$, and Phone B starts at $40\%$. 2) After $30$ minutes; both are at $55\%$. 3) $85\%$ 4) $a(x)=1.5x+10$, for $0\le x\le60$; $b(x)=0.5x+40$, for $0\le x\le100$

Interpret rate of change and initial value 93

The graph shows the amount of water in tanks A and B over $8$ minutes. The curve labeled $a$ represents Tank A, and the curve labeled $b$ represents Tank B. a) How much water was in each tank at $t=0$? b) One tank is being filled and the other is being drained. Find the rate of change for each tank during its linear filling or draining phase, in liters per minute. c) After how many minutes do the tanks contain the same amount of water? What is that amount?

Hints

- Read each initial value where $t=0$. - Find each rate using change in volume divided by change in time. - Equal amounts occur where the two graph curves intersect.

Solution

1. At $t=0$, Tank A contains $5\,\text{L}$, and Tank B contains $35\,\text{L}$. 2. Tank A rises from $5\,\text{L}$ to $15\,\text{L}$ in $4$ minutes. Its rate is $\frac{15-5}{4-0}=2.5\,\text{L/min}$. 3. Tank B falls from $35\,\text{L}$ to $15\,\text{L}$ in $4$ minutes. Its rate of change is $\frac{15-35}{4-0}=-5\,\text{L/min}$, so it is drained at $5\,\text{L/min}$. 4. The graphs intersect at $(4, 15)$. After $4$ minutes, both tanks contain $15\,\text{L}$.

Answer

a) Tank A: $5\,\text{L}$; Tank B: $35\,\text{L}$ b) Tank A: $2.5\,\text{L/min}$; Tank B: $-5\,\text{L/min}$, or draining at $5\,\text{L/min}$ c) After $4$ minutes; $15\,\text{L}$

Interpret rate of change and initial value 94

The equation $x-2y=4$ defines a linear function. a) Rewrite the equation in the form $y=mx+b$. State the rate of change $m$ and the initial value $b$. b) Which graph represents the function? Justify your choice using both the rate of change and the initial value.

Hints

- Put the equation into slope-intercept form before looking at the choices. - In $y=mx+b$, identify separately what $m$ and $b$ tell you about the graph. - A correct choice must match both the direction/steepness and the y-axis crossing.

Solution

1. Solve for $y$: $x-2y=4$ gives $-2y=4-x$, so $y=\frac{1}{2}x-2$. 2. The rate of change is $m=\frac{1}{2}$, and the initial value is $b=-2$. 3. Graph a) rises from left to right with slope $\frac{1}{2}$ and crosses the y-axis at $-2$. The other graphs have either the wrong slope sign or the wrong initial value.

Answer

a) $y=\frac{1}{2}x-2$; $m=\frac{1}{2}$, $b=-2$ b) a)

Interpret rate of change and initial value 95

A rain barrel holds $68$ gallons at $3{:}00$ p.m. and $52$ gallons at $7{:}00$ p.m. It has been leaking at a constant rate since noon and keeps leaking at that rate until empty. Let $t$ be hours since noon and $u$ be hours since $3{:}00$ p.m. a) How many gallons does the barrel lose each hour? b) Write a rule $V(t)$ for the water left using $t$, and a rule $W(u)$ using $u$. c) Explain why the two rules have different starting values even though they describe the same barrel. d) At what clock time will the barrel be empty?

Hints

- Compare the two readings and the number of hours between them. - For each rule, decide what its input of zero means. - What output means the barrel is empty?

Solution

1. From $3{:}00$ p.m. to $7{:}00$ p.m., the barrel loses $68-52=16$ gallons in $4$ hours, so the rate is $-4$ gallons per hour. 2. Noon is $3$ hours before the first reading, so there were $68+3\cdot4=80$ gallons at noon. Thus $V(t)=80-4t$, valid for $0\le t\le20$. 3. At $3{:}00$ p.m. there were $68$ gallons, so $W(u)=68-4u$, valid for $0\le u\le17$. 4. The rules have the same rate but different starting values because $t=0$ means noon and $u=0$ means $3{:}00$ p.m. 5. The barrel reaches zero after $20$ hours from noon, which is $8{:}00$ a.m. the next day.

Answer

a) $4$ gallons lost per hour; rate $-4$ gallons per hour. b) $V(t)=80-4t$, for $0\le t\le20$; $W(u)=68-4u$, for $0\le u\le17$. c) The starting values refer to different times: noon and $3{:}00$ p.m. d) $8{:}00$ a.m. the next day.

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Interpret rate of change and initial value

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