- Compare consecutive output changes for equal input changes. - A linear function has a constant rate of change.

1. Each time \(x\) increases by \(1\), the output increases by \(3\): \(7-4=3\), \(10-7=3\), and \(13-10=3\). 2. A constant rate of change means the points lie on a straight line, so the function is linear.

Linear. The output changes by a constant \(3\) for every increase of \(1\) in \(x\).

- Compare the spacing of the inputs. - Find the changes between consecutive outputs and check whether they stay the same.

1. Every time the input increases by \(1\), the output increases by \(2\). 2. The rate of change is constant, so the relationship is linear.

Yes. The output increases by \(2\) for each increase of \(1\) in the input.

- Think about how you recognize a linear graph. - Compare outputs at two different inputs.

1. The graph is a straight horizontal line, so it represents a linear function. 2. The output changes by \(0\) for every input increase, so the rate is \(0\).

Linear. The output stays constant, so its rate of change is \(0\).

- Imagine writing the next three weekly gains for each plant. - Compare the gains over individual weeks, rather than the plants' total heights.

1. The first plant gains the same amount over equal one-week intervals, so its rate is constant. 2. The second plant has changing weekly gains, so it does not have a constant rate over those weeks.

The first plant. Equal weekly height gains give a constant rate of change.

- Write a rule for each relationship. - Check whether equal increases in the input produce equal increases in the output. - Compare what happens when the input is doubled.

1. The taxi relationship is linear. It has a fixed initial value of \(\$3.50\) and a constant rate of \(\$1.80\) per mile, so \(C(x)=1.80x+3.50\). 2. The patio relationship is not linear. Its area is \(A(s)=s^2\). For example, doubling the side length multiplies the area by \(4\), not by \(2\).

1) Linear, because the cost increases at a constant rate from a fixed starting value. 2) Not linear, because \(A=s^2\).

- A linear function changes by the same amount per unit of input. - The two input intervals have different lengths; compare change in output per unit of input. - Do the two rates match?

1. From input \(-2\) to \(2\), the rate of change is \(\frac{3-5}{2-(-2)}=-\frac{1}{2}\). 2. From input \(2\) to \(5\), the rate of change is \(\frac{1-3}{5-2}=-\frac{2}{3}\). 3. The rates differ, so these values cannot all lie on the graph of a single linear function.

No. The two rates of change are \(-\frac{1}{2}\) and \(-\frac{2}{3}\), so the rate is not constant.

- What must be true about the graph of a proportional relationship? - Consider the role of \(b\) in \(y=mx+b\). - What output does a proportional function have when \(x=0\)?

1. A linear function has the form \(y=mx+b\). A proportional function has the form \(y=kx\), which is the special case \(b=0\). 2. Therefore, the statement is false. Only linear functions whose graphs pass through the origin are proportional. 3. For example, \(y=2x+5\) is linear but not proportional because its y-intercept is \(5\).

The statement is false. A linear function \(y=mx+b\) is proportional only when \(b=0\). For example, \(y=2x+5\) is linear but not proportional.

- Identify each year as an input and its charging-port count as the corresponding output. - Choose a vertical scale large enough to include the greatest value. - For the increase, compare the two years named in the question. - To test for a constant yearly increase, compare each value with the one immediately before it.

1. The points are \((2017, 10700)\), \((2018, 15600)\), \((2019, 23900)\), \((2020, 39500)\), and \((2021, 52200)\). A vertical scale from \(0\) to about \(55{,}000\) is appropriate. 2. The increase from \(2019\) to \(2020\) is \(39{,}500-23{,}900=15{,}600\). 3. The yearly first differences are \(4900\), \(8300\), \(15{,}600\), and \(12{,}700\). Because these differences are not equal, the data do not show a constant yearly increase.

a) \((2017, 10700)\), \((2018, 15600)\), \((2019, 23900)\), \((2020, 39500)\), and \((2021, 52200)\); a vertical scale from \(0\) to about \(55{,}000\) b) \(15{,}600\) charging ports c) No. The first differences are not constant.

- Substitute each speed into the model or read the corresponding graph value. - For part b, compare the second stopping distance with twice the first and form the requested ratio. - For part c, use the shape of the graph or compare the formula with the general form of a linear function; do not rely only on the doubling test.

1. At \(30\,\text{mph}\), \(s(30)=30+\frac{30^2}{20}=30+45=75\,\text{ft}\). 2. At \(60\,\text{mph}\), \(s(60)=60+\frac{60^2}{20}=60+180=240\,\text{ft}\). 3. Twice the first distance would be \(150\,\text{ft}\), not \(240\,\text{ft}\). The ratio is \(\frac{240}{75}=3.2\), so doubling speed multiplies the stopping distance by \(3.2\), not by \(2\). 4. The function is nonlinear. Its formula contains a \(v^2\) term rather than having the form \(mv+b\), and its graph is curved rather than a straight line. The doubling result in part b by itself would not prove nonlinearity, because a linear function with a nonzero initial value also need not double its output when the input doubles.

a) \(75\,\text{ft}\) at \(30\,\text{mph}\); \(240\,\text{ft}\) at \(60\,\text{mph}\) b) No. The ratio is \(3.2\). c) Nonlinear; the formula contains a \(v^2\) term and the graph is curved rather than a straight line.

- Look at whether the dots line up as the side length grows. - Think about what a graph of a linear function looks like.

1. The plotted points turn upward more sharply as the side length grows; they cannot all lie on one straight line. 2. A linear function would have its points on a single straight line, so this relationship is nonlinear.

Nonlinear. The plotted points do not lie on a single straight line.

- Compare the changes in \(y\) over equal changes in \(x\). - Ask what must stay constant for a relationship to be linear.

1. A linear relationship has a constant rate of change, so equal changes in \(x\) produce equal changes in \(y\). 2. Here the changes in \(y\) are \(2,4,6,8\), which are not constant. 3. Therefore, the relationship is nonlinear.

The relationship is nonlinear because equal changes in \(x\) do not produce a constant change in \(y\).

- Compare the sizes of the input steps before judging the output differences. - Could unequal input steps explain unequal output changes?

1. From input \(0\) to \(2\), the rate is \((7-1)\div(2-0)=3\). 2. From input \(2\) to \(5\), the rate is \((16-7)\div(5-2)=3\). 3. The rate is constant, so the data are consistent with a linear function; unequal output differences correspond to unequal input intervals.

No. Both rates are \(3\) per input unit, so the data fit a linear function.

- Compare the rise as you move one square to the right in different parts of each graph. - Does either graph have a rate that changes from one part to another?

1. Panel a) is a straight line, so its rate is constant and the function is linear. 2. Panel b) bends upward. Its slope becomes steeper as the input increases, so its rate changes and it is nonlinear.

Panel a) is linear. Panel b) is nonlinear because the curve gets steeper.

- Compare the height increase during each equal time interval. - With constant inflow, decide whether a wider container makes the water level rise faster or more slowly.

1. In container A, the height increases by exactly \(2.0\,\text{cm}\) every \(10\) seconds. With a constant inflow, this constant rate of height increase indicates a uniform cross section, like a cylinder. 2. The graph for A is a line through the origin. The graph for B increases but becomes less steep. 3. The successive height increases for B are \(3.5\), \(2.5\), \(1.5\), \(1.0\), and \(0.5\) centimeters. Because the same volume produces a smaller rise as the water gets higher, container B becomes wider toward the top.

a) Container A, because its height increases by a constant \(2.0\,\text{cm}\) every \(10\) seconds. b) A is linear; B is increasing and becomes less steep. c) Container B becomes wider toward the top because the height increases by progressively smaller amounts.

- Multiply each quantity by the unit price that applies to that order size. - Compare the total costs for \(9\), \(10\), and \(11\) shirts carefully around the pricing break. - A proportional relation has one constant unit rate, while a linear relation has one constant rate of change across its full domain. - Think about which input values are actually possible when the input counts shirts.

1. Ten shirts cost \(10(\$15.00)=\$150.00\). Eleven shirts cost \(11(\$12.00)=\$132.00\). Therefore, buying \(11\) shirts costs \(\$18.00\) less than buying \(10\). 2. Nine shirts cost \(9(\$15.00)=\$135.00\). Eleven shirts cost \(\$132.00\), so the customer would receive two additional shirts and spend \(\$3.00\) less. 3. For whole-number quantities, \(C(n)=15n\) for \(1\le n\le10\), and \(C(n)=12n\) for \(n\ge11\). The table is: <table><tr><td>Shirts</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><td>\(9\)</td><td>\(10\)</td><td>\(11\)</td><td>\(12\)</td><td>\(13\)</td><td>\(14\)</td><td>\(15\)</td></tr><tr><td>Total price</td><td>\(\$15.00\)</td><td>\(\$30.00\)</td><td>\(\$45.00\)</td><td>\(\$60.00\)</td><td>\(\$75.00\)</td><td>\(\$90.00\)</td><td>\(\$105.00\)</td><td>\(\$120.00\)</td><td>\(\$135.00\)</td><td>\(\$150.00\)</td><td>\(\$132.00\)</td><td>\(\$144.00\)</td><td>\(\$156.00\)</td><td>\(\$168.00\)</td><td>\(\$180.00\)</td></tr></table> The unit price is not constant across the full domain, so the relation is not proportional. It is not linear over its full domain because the change in total price is not constant; for example, the total drops by \(\$18.00\) from \(10\) to \(11\) shirts instead of continuing at one constant rate. The relation is discrete because fractional shirts are not ordered.

a) \(10\) shirts cost \(\$150.00\); \(11\) shirts cost \(\$132.00\). The larger order costs \(\$18.00\) less. b) \(9\) shirts cost \(\$135.00\), so ordering \(11\) shirts for \(\$132.00\) saves \(\$3.00\). c) <table><tr><td>Shirts</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><td>\(9\)</td><td>\(10\)</td><td>\(11\)</td><td>\(12\)</td><td>\(13\)</td><td>\(14\)</td><td>\(15\)</td></tr><tr><td>Total price</td><td>\(\$15.00\)</td><td>\(\$30.00\)</td><td>\(\$45.00\)</td><td>\(\$60.00\)</td><td>\(\$75.00\)</td><td>\(\$90.00\)</td><td>\(\$105.00\)</td><td>\(\$120.00\)</td><td>\(\$135.00\)</td><td>\(\$150.00\)</td><td>\(\$132.00\)</td><td>\(\$144.00\)</td><td>\(\$156.00\)</td><td>\(\$168.00\)</td><td>\(\$180.00\)</td></tr></table> The relation is neither proportional nor linear over its full domain because it does not have one constant rate of change. Its graph is discrete because only whole-number quantities of shirts are allowed.

- For a linear function, output changes at a constant rate per input unit. - The input gaps here differ, so divide the output change by the matching input gap. - Compare the two rates.

1. From \(x=1.5\) to \(x=4\), the rate is \(\frac{9-4}{4-1.5}=\frac{5}{2.5}=2\). 2. From \(x=4\) to \(x=6\), the rate is \(\frac{13.5-9}{6-4}=\frac{4.5}{2}=2.25\). 3. The rates differ, so the values do not come from one linear function.

No. The two rates of change are \(2\) and \(2.25\), which are not equal.

- Compare output change per input unit, not just changes in output. - If the rate stays constant, use one known input-output pair to find the value at input \(0\). - To find the last input, work backward from its given output.

1. The rate from \((1,5)\) to \((2,8)\) is \(\frac{8-5}{2-1}=3\). From \((2,8)\) to \((4,14)\), it is \(\frac{14-8}{4-2}=3\). The rate is constant. 2. With rate \(3\), the function has form \(f(x)=3x+b\). Since \(f(1)=5\), \(b=2\), so \(f(x)=3x+2\). 3. Thus \(f(0)=2\). Solving \(3x+2=26\) gives \(x=8\).

The table is linear, with rule \(f(x)=3x+2\). <table><tr><th>\(x\)</th><td>\(1\)</td><td>\(2\)</td><td>\(4\)</td><td>\(0\)</td><td>\(8\)</td></tr><tr><th>\(f(x)\)</th><td>\(5\)</td><td>\(8\)</td><td>\(14\)</td><td>\(2\)</td><td>\(26\)</td></tr></table>

- Substitute the same three inputs into each rule. - For equally spaced inputs, a linear function has equal output differences. - A proportional rule must have output \(0\) at input \(0\), as well as a constant rate.

1. For a), the outputs are \(0\), \(0.8\), and \(1.6\). Each step adds \(0.8\), and the rule begins at \(0\). It is linear and proportional. 2. For b), the outputs are \(2\), \(2.8\), and \(3.6\). Each step adds \(0.8\), but the initial value is \(2\). It is linear but not proportional. 3. For c), the outputs are \(0\), \(1\), and \(4\). The increases are \(1\) and \(3\), so it is nonlinear and not proportional.

a) Outputs \(0,0.8,1.6\); linear and proportional. b) Outputs \(2,2.8,3.6\); linear but not proportional. c) Outputs \(0,1,4\); nonlinear and not proportional.

- Compare average changes over more than one interval of time. - Divide temperature change by elapsed minutes for each interval. - Think about what the table tells you at observed times versus times not measured.

1. From \(0\) to \(2\) minutes, the average change is \(\frac{140-170}{2}=-15\,^{\circ}\text{F}\) per minute. From \(2\) to \(4\), it is \(\frac{118-140}{2}=-11\,^{\circ}\text{F}\) per minute. The rates differ, so the data do not show a linear relationship. 2. The cocoa cools faster at first and then more slowly. For example, from \(10\) to \(15\) minutes the average change is \(\frac{74-84}{5}=-2\,^{\circ}\text{F}\) per minute. 3. No measurement or temperature rule is supplied for \(t=30\), so the table cannot determine its exact temperature then.

a) No; the average rates of change are not constant. b) The drop per minute becomes smaller in magnitude as time passes. c) No. The table ends at \(15\) minutes and gives no rule for later temperatures.

- What must be true about the ratios \(\frac{y}{x}\) in a proportional relationship? - What point must every proportional relationship contain? - How can you use changes in \(y\) and changes in \(x\) to test for a constant rate of change?

1. For Table 1, check the ratios \(\frac{y}{x}\): \(\frac{2}{1.5}=\frac{4}{3}\), \(\frac{4}{3}=\frac{4}{3}\), \(\frac{6}{4.5}=\frac{4}{3}\), and \(\frac{8}{6}=\frac{4}{3}\). The ratio is constant, so the relationship is proportional. Every proportional relationship is also linear. 2. Table 2 contains the point \((0, 3)\), so it is not proportional because a proportional relationship must contain \((0, 0)\). 3. For Table 2, calculate the rate of change between consecutive points: \(\frac{3-(-1)}{0-(-2)}=2\), \(\frac{7-3}{2-0}=2\), and \(\frac{11-7}{4-2}=2\). The rate of change is constant, so the relationship is linear.

Table 1 is proportional and linear because every ratio \(\frac{y}{x}\) equals \(\frac{4}{3}\). Table 2 is linear because its rate of change is always \(2\), but it is not proportional because it contains \((0, 3)\) rather than \((0, 0)\).

- How can you compare the slopes between pairs of points? - Once you know the slope, how can you find the y-intercept? - What must the y-intercept be for a proportional relationship?

1. Compare slopes between consecutive points: \(\frac{5-11}{-1-(-4)}=-2\), \(\frac{-1-5}{2-(-1)}=-2\), and \(\frac{-7-(-1)}{5-2}=-2\). Since the rate of change is constant, the relationship is linear. 2. Use slope \(-2\) and point \((2, -1)\): \(-1=-2\cdot2+b\), so \(b=3\). Thus \(y=-2x+3\). 3. A proportional relationship has the form \(y=kx\) and therefore has y-intercept \(0\). Here the y-intercept is \(3\), so the relationship is not proportional.

a) The slope between every pair of consecutive points is \(-2\), so the function is linear. b) \(y=-2x+3\) c) No. The y-intercept is \(3\), not \(0\), so the relationship is not proportional.

- Compare how far apart the first two inputs and outputs are. - For part a), think about what must stay the same from one input interval to the next. - Test the proposed output in part b) against the change you found in part a).

1. From input \(-2\) to \(1\), the rate is \(\frac{1-7}{1-(-2)}=-2\) per input unit. 2. From \(1\) to \(4\), the input also increases by \(3\). A linear pattern requires the output to decrease by \(6\), giving \(1-6=-5\). 3. An output of \(-4\) would give a change of \(-5\) over that interval, a different rate, so it would not make the pattern linear.

a) \(-5\), because the rate must remain \(-2\) per input unit. b) No. An output of \(-4\) would give a different rate of change.

- First decide which displayed graph is a straight line through the origin, which is a straight line with a nonzero y-intercept, and which is curved. - A proportional linear function has a straight graph through the origin; another straight graph can still be linear without being proportional. - Test one clearly readable point from each graph in the candidate equations to confirm the match.

1. Graph a) is Equation (3), \(g(x)=-1.5x\). The point \((2, -3)\) lies on the graph because \(g(2)=-1.5\cdot2=-3\). The graph is a line through the origin, so it is proportional. 2. Graph b) is Equation (2), \(f(x)=2x+1\). The point \((0, 1)\) lies on the graph because \(f(0)=1\). The graph is a line that does not pass through the origin, so it is linear but not proportional. 3. Graph c) is Equation (1), \(h(x)=x^2\). The point \((2, 4)\) lies on the graph because \(h(2)=4\). The graph is curved, so it is nonlinear.

a) Equation (3); proportional; example point \((2, -3)\) b) Equation (2); linear but not proportional; example point \((0, 1)\) c) Equation (1); nonlinear; example point \((2, 4)\)

- Connect cross-sectional area with how quickly water height changes. - A straight line represents a constant rate of change. - In a container that widens, each additional unit of height requires more water.

1. A cylinder has the same cross-sectional area at every height. With a constant inflow, its water height rises at a constant rate, producing a linear graph. Therefore, Graph \(f\) represents Container A. 2. Container B has a larger cross-sectional area higher up. Each additional unit of height requires more water than the previous one, so the water height rises more slowly over time. The graph therefore flattens, as Graph \(g\) does.

a) Graph \(f\): Container A; Graph \(g\): Container B b) The cylinder's height increases at a constant rate, while the widening container's height increases at a decreasing rate.

- What special point must a proportional graph contain? - Look at where the graph crosses the y-axis. - Choose two points and compare the horizontal and vertical changes. - Check whether the same rate of change holds elsewhere on the line.

1. A proportional relationship must graph as a line through \((0, 0)\). This graph crosses the y-axis at \(2\), so Maya is incorrect. 2. Using \((0, 2)\) and \((2, 3)\), the change in \(y\) is \(1\) when the change in \(x\) is \(2\). The slope is \(\frac{1}{2}=0.5\), and it is constant along the line. Therefore, Jordan is correct.

Maya is incorrect because the line does not pass through \((0, 0)\). Jordan is correct because the constant slope is \(0.5\), so every increase of \(2\) in \(x\) produces an increase of \(1\) in \(y\).

- A linear function graphs as a line; a quadratic function graphs as a parabola. - Intersections represent equal heights. - Evaluate both functions at \(t=10\) and subtract.

1. The function \(h_1(t)=15t\) is linear, so it matches the straight line. The function \(h_2(t)=t^2\) is quadratic, so it matches the parabola. 2. The graphs intersect at \(t=0\) and \(t=15\). Substitution verifies both times: \(h_1(0)=h_2(0)=0\), and \(h_1(15)=15\cdot15=225\) while \(h_2(15)=15^2=225\). After the start, the heights are equal again at \(15\) minutes, when both are \(225\,\text{cm}\). 3. \(h_1(10)=15\cdot10=150\,\text{cm}\) and \(h_2(10)=10^2=100\,\text{cm}\). The difference is \(50\,\text{cm}\).

a) Tank 1: straight line; Tank 2: parabola b) \(t=0\) and \(t=15\) minutes; after the start, \(15\) minutes at \(225\,\text{cm}\) c) \(50\,\text{cm}\)

- Write each area formula and substitute the fixed height. - To test proportionality, look for a constant multiple of \(r\) with no added constant. - To classify linearity, compare each formula with the form \(mr+b\); remember that a relationship can be linear without being proportional.

1. The lateral surface area is \(L(r)=2\pi rh\). With \(h=10\), \(L(r)=20\pi r\). This has the form \(mr\) with a constant rate, so it is proportional and linear. 2. The base area is \(B(r)=\pi r^2\). The variable is squared, so the relationship does not have the form \(mr+b\). It is not proportional and is nonlinear. For example, doubling \(r\) multiplies \(B\) by \(4\), but the nonlinear classification follows from the squared variable rather than from nonproportionality alone.

a) Proportional and linear, because \(L(r)=20\pi r\). b) Not proportional and nonlinear, because \(B(r)=\pi r^2\).

- A straight line represents a constant rate of change in height. - A wider horizontal layer contains more water and therefore takes longer to drain through. - Compare how each graph's steepness changes over time.

1. Graph \(h_2\) decreases slowly at first and then becomes increasingly steep, so the water height decreases faster over time. Graph \(h_3\) decreases steeply at first and then flattens, so the water height decreases more slowly over time. 2. Graph \(h_1\) matches Container 1 because a cylinder has constant cross-sectional area, producing a constant rate of decrease in height. Graph \(h_2\) matches Container 2 because the wide upper portion makes the height drop slowly at first, while the narrower lower portion makes it drop faster later. Graph \(h_3\) matches Container 3 because the narrow upper portion makes the height drop quickly at first, while the wider lower portion makes it drop more slowly later.

a) \(h_2\) decreases slowly and then faster; \(h_3\) decreases quickly and then more slowly. b) \(h_1\) matches 1; \(h_2\) matches 2; \(h_3\) matches 3.