- The initial value of a function is its output when the input is \(0\). - Read the \(x=0\) row entry for the table and substitute \(0\) into the equation.

1. The initial value is the output when \(x=0\). 2. From the table, \(f(0)=3\). 3. From the equation, \(g(0)=2(0)+1=1\). 4. Because \(3>1\), function \(f\) has the greater initial value.

\(f\) has the greater initial value. The initial values are \(f(0)=3\) and \(g(0)=1\).

- The input for comparison is zero for both functions. - Use the equation for one output and the table for the other.

1. \(A(0)=2(0)+3=3\). 2. The table shows \(B(0)=5\), and \(5>3\), so \(B\) has the greater initial value.

Function \(B\), with \(5\) compared with \(3\) for \(A\).

- Count the rise for a one-unit change of input on the graph. - Find the output difference between adjacent table columns. - Compare the changes, not the starting values.

1. On the graph, \(A\) rises from \(0\) to \(6\) as \(x\) rises from \(0\) to \(3\), so its rate is \(2\). 2. In the table, \(B\) rises by \(3\) whenever \(x\) rises by \(1\), so its rate is \(3\). 3. Function \(B\) has the greater rate.

\(B\). Its rate is \(3\) per input unit, compared with \(2\) for \(A\).

- Find the cost for the requested number at each shop. - Use multiplication for the shop that gives a price per notebook.

1. At the first shop, three notebooks cost \(3\cdot\$4=\$12\). 2. The table gives \(\$13\) for three notebooks at the second shop, so the first shop costs less.

The first shop: \(\$12\), compared with \(\$13\).

- Look at how many stickers each person adds during one week. - Use the table to find Nico's weekly increase and starting amount.

1. Mia adds \(3\) stickers each week. 2. Nico adds \(8-5=3\) stickers in the first week and \(11-8=3\) in the next. 3. Nico starts with \(5\) stickers, compared with Mia's \(2\).

Yes, both add \(3\) stickers per week. Nico starts with more: \(5\) stickers versus Mia's \(2\).

- On the graph, use two clear points to find change in output divided by change in input. - In the table, compare output changes with the corresponding input changes. - The initial value is the output at \(x=0\) in each representation.

1. From the graph, \(f\) passes through \((0,2)\) and \((4,10)\), so its rate of change is \(\frac{10-2}{4-0}=2\). Its initial value is \(2\). 2. From the table, \(g\) changes by \(4\) when \(x\) changes by \(2\), so its rate of change is \(\frac{4}{2}=2\). Its initial value is \(g(0)=5\). 3. The functions have the same rate of change, but \(g\) has the greater initial value by \(5-2=3\).

The rates are equal at \(2\) output units per input unit. The initial values are \(2\) for \(f\) and \(5\) for \(g\), so \(g\) has the greater initial value by \(3\).

- Read the graph's height at the specified input. - Substitute the same input into the equation. - The function with the larger output depends on which input is used.

1. The graph gives \(f(1)=5\), and \(g(1)=3\). Thus \(f\) is greater at \(1\). 2. The graph gives \(f(4)=8\), and \(g(4)=9\). Thus \(g\) is greater at \(4\).

a) \(f\): \(5\) versus \(3\) b) \(g\): \(9\) versus \(8\)

- A rate compares change in the output with change in the input. - The table's time gaps are not one minute, so account for their length. - Compare the rates, not the total gallons already recorded.

1. The first hose has rate \(5\) gallons per minute. 2. For the second hose, the volume rises \(8\) gallons in \(2\) minutes, giving \(4\) gallons per minute. 3. The first hose's rate is \(1\) gallon per minute greater.

The first hose, by \(1\) gallon per minute.

- Compare the scores after the first requested round before considering the later one. - Match the round number with the dot directly above it. - Look at how the gap between the scores changes as rounds pass.

1. \(A(1)=1+3=4\), while the graph shows \(B(1)=8\), so \(B\) is greater. 2. \(A(5)=1+3\cdot5=16\), while \(B(5)=12\), so \(A\) is greater. 3. \(A\) gains \(3\) points per round, faster than the graph's rate of \(1\) point per round for \(B\). Its lower start is eventually overcome.

a) \(B\): \(8\) points compared with \(4\) b) \(A\): \(16\) points compared with \(12\) c) \(A\) starts lower but increases faster each round.

- What would it take for two functions to agree on every listed input? - Try another input from the table and compare its two outputs.

1. At \(x=2\), \(f(2)=5\) and \(g(2)=5\). 2. At \(x=0\), \(f(0)=3\) but \(g(0)=5\). The functions disagree at an input, so they are not the same function.

No. For example, \(f(0)=3\) but \(g(0)=5\).

- Test consecutive whole-number movie counts near the flat monthly price. - Compare each result with \(\$9.99\). - At what movie count does Service B first exceed the flat price?

1. For \(4\) movies, Service B costs \(\$1.50 \cdot 4=\$6.00\). For \(8\) movies, it costs \(\$1.50 \cdot 8=\$12.00\). 2. At \(6\) movies, Service B costs \(\$9.00\), which is less than \(\$9.99\). At \(7\) movies, Service B costs \(\$10.50\), which is more than \(\$9.99\). Therefore, Service A is less expensive starting at \(7\) movies.

a) \(4\) movies cost \(\$6.00\); \(8\) movies cost \(\$12.00\). b) Service A is less expensive starting at \(7\) movies per month.

- Evaluate each formula separately. - Find an absolute difference when comparing the two outputs. - Set the expressions equal to locate the boundary value. - Compare their rates of change on either side of that value.

1. At \(30\,^{\circ}\text{C}\), the estimate gives \(2 \cdot 30 + 30 = 90\), while the exact formula gives \(1.8 \cdot 30 + 32 = 86\). 2. At \(25\,^{\circ}\text{C}\), the estimate gives \(2 \cdot 25 + 30 = 80\), and the exact formula gives \(1.8 \cdot 25 + 32 = 77\). The difference is \(3\,^{\circ}\text{F}\). 3. Set the two expressions equal to find their intersection: \(2C + 30 = 1.8C + 32\). 4. Subtract \(1.8C + 30\): \(0.2C = 2\), so \(C = 10\). 5. At \(10\,^{\circ}\text{C}\), both formulas give \(50\,^{\circ}\text{F}\). Since the estimate has the greater rate of change, it is greater only when \(C > 10\). The claim is false because it includes \(C = 10\).

a) Estimate: \(90\,^{\circ}\text{F}\); exact value: \(86\,^{\circ}\text{F}\) b) The estimate is \(3\,^{\circ}\text{F}\) higher. c) The claim is false. The values are equal at \(10\,^{\circ}\text{C}\), and the estimate is higher only for \(C > 10\).

- Interpret the coefficient in Paint B's equation. - Find Paint A's coverage for one gallon. - Use the better unit rate to find the gallons needed for the target area.

1. Paint A covers \(425\div 5=85\) square feet per gallon. 2. In \(y=90x\), the coefficient means Paint B covers \(90\) square feet per gallon. 3. Since \(90>85\), Paint B has greater coverage. 4. The amount needed is \(630\div 90=7\) gallons.

a) Paint B; it covers \(90\,\text{ft}^2\) per gallon, compared with \(85\,\text{ft}^2\) for Paint A. b) \(7\) gallons.

- How do you find slope from two points in a table? - What does slope tell you when comparing two linear functions? - At an intersection point, what must be true about the two function values?

1. From the table, the slope of \(f\) is \(m_f = \frac{8 - 2}{3 - 1} = \frac{6}{2} = 3\). 2. Use \((1, 2)\) in \(f(x) = 3x + b\): \(2 = 3 \cdot 1 + b\), so \(b = -1\). Thus \(f(x) = 3x - 1\). 3. The slopes are \(3\) for \(f\) and \(4\) for \(g\), so \(g\) has the greater slope. 4. Set the functions equal: \(3x - 1 = 4x - 1\). Then \(x = 0\). 5. Substituting gives \(y = -1\), so the intersection point is \((0, -1)\).

a) \(f(x) = 3x - 1\) b) \(g\) has the greater slope. c) \((0, -1)\)

- How are speed, time, and distance related? - What does the coefficient in \(y=kx\) represent? - Find Tim's rate from one distance-time pair.

1. Tim's speed is \(12.5\div 5=2.5\) meters per second; the second pair confirms \(30\div 12=2.5\). 2. Sarah's speed is the coefficient \(2.4\) meters per second. Since \(2.5>2.4\), Tim is faster. 3. Tim runs \(2.5\cdot 20=50\) meters. 4. For Sarah, solve \(60=2.4t\): \(t=60\div 2.4=25\) seconds.

a) Tim is faster: \(2.5\,\text{m/s}\) compared with Sarah's \(2.4\,\text{m/s}\). b) \(50\) meters. c) \(25\) seconds.

- How do you find slope from two points? - Translate the verbal description of Liquid 2 into a rate and an initial value. - Which number in \(mt + b\) represents the rate, and which represents the starting value? - Substitute \(t = 10\) into all three functions for the final comparison.

1. Liquid 1 is already given: \(T_1(t) = 4t + 68\). 2. Liquid 2 has initial value \(65\) and rate \(5\), so \(T_2(t) = 5t + 65\). 3. For Liquid 3, the slope is \(m = \frac{83 - 74}{5 - 2} = 3\). Using \((2, 74)\), \(74 = 3 \cdot 2 + b\), so \(b = 68\). Thus \(T_3(t) = 3t + 68\). 4. The initial temperatures are \(68^\circ\text{F}\), \(65^\circ\text{F}\), and \(68^\circ\text{F}\), so Liquids 1 and 3 tie for the highest initial temperature. 5. The rates are \(4\), \(5\), and \(3\) degrees Fahrenheit per minute, so Liquid 2 warms fastest. 6. At \(t = 10\), the temperatures are \(T_1(10) = 108^\circ\text{F}\), \(T_2(10) = 115^\circ\text{F}\), and \(T_3(10) = 98^\circ\text{F}\).

a) \(T_1(t) = 4t + 68\); \(T_2(t) = 5t + 65\); \(T_3(t) = 3t + 68\) b) Liquids 1 and 3, at \(68^\circ\text{F}\) c) Liquid 2, at \(5^\circ\text{F}\) per minute d) After \(10\) minutes: Liquid 1 is \(108^\circ\text{F}\), Liquid 2 is \(115^\circ\text{F}\), and Liquid 3 is \(98^\circ\text{F}\).

- What does the coefficient of \(x\) represent in \(y=4.80x\)? - Find Variety A's price for one pound. - You can compare the total prices or first find the difference in price per pound.

1. The unit price for Variety A is \(13.50\div 3=\$4.50\) per pound. 2. From \(y=4.80x\), the unit price for Variety B is \(\$4.80\) per pound. 3. Variety B is more expensive because \(4.80>4.50\). 4. The difference per pound is \(\$4.80-\$4.50=\$0.30\). For \(2.5\,\text{lb}\), the difference is \(0.30\cdot 2.5=\$0.75\).

a) Variety B costs more: \(\$4.80\) per pound compared with \(\$4.50\) per pound for Variety A. b) Variety B costs \(\$0.75\) more for \(2.5\,\text{lb}\).

- Use slope-intercept form to write \(g\). - Compare the absolute values of the slopes to compare steepness. - At an intersection, both functions have the same output for the same input.

1. From the given slope and y-intercept, \(g(x) = 0.5x - 1\). 2. Steepness is determined by the absolute value of the slope. Since \(|-2| = 2\) and \(|0.5| = 0.5\), the graph of \(f\) is steeper. 3. At the intersection, \(-2x + 4 = 0.5x - 1\). Thus, \(5 = 2.5x\), so \(x = 2\). Then \(y = 0\), giving \(S(2, 0)\).

a) \(g(x) = 0.5x - 1\) b) \(f\) is steeper because \(|-2| > |0.5|\). c) \(S(2, 0)\)

- Use the stated constant-rate condition to determine Panel A's energy production per hour. - Compare the two panels using energy produced per hour. - For part c, first determine how much energy Panel B produces in \(6\) hours.

1. Panel A's rate is \(1.2\div5=0.24\,\text{kWh/h}\), so \(E_A(t)=0.24t\). 2. In \(8\) hours, Panel A produces \(0.24\cdot8=1.92\,\text{kWh}\). 3. Panel B produces more per hour because \(0.25>0.24\). 4. Panel B produces \(0.25\cdot6=1.5\,\text{kWh}\) in \(6\) hours. Solve \(1.5=0.24t\): \(t=1.5\div0.24=6.25\) hours, or \(6\) hours \(15\) minutes.

a) \(1.92\,\text{kWh}\) b) Panel B, because \(0.25\,\text{kWh/h}>0.24\,\text{kWh/h}\) c) \(6.25\) hours, or \(6\) hours \(15\) minutes

- What does the coefficient of \(t\) tell you about Beta's rate? - How much does Alpha print in one hour? - Find each printer's time for the same volume. - Convert the difference in hours to minutes.

1. Alpha's rate is \(50 \div 2=25\,\text{cm}^3/\text{h}\). Beta's rate is \(22.5\,\text{cm}^3/\text{h}\). Since \(25>22.5\), Alpha is faster. 2. Alpha prints \(25\), \(50\), \(75\), and \(100\,\text{cm}^3\) in \(1\), \(2\), \(3\), and \(4\) hours. 3. Alpha needs \(180 \div 25=7.2\) hours. Beta needs \(180 \div 22.5=8\) hours. The difference is \(0.8\) hour, or \(0.8 \cdot 60=48\) minutes.

a) Alpha is faster: \(25\,\text{cm}^3/\text{h}\) compared with \(22.5\,\text{cm}^3/\text{h}\). b) <table> <tr><th>Time</th><td>\(1\,\text{h}\)</td><td>\(2\,\text{h}\)</td><td>\(3\,\text{h}\)</td><td>\(4\,\text{h}\)</td></tr> <tr><th>Volume</th><td>\(25\,\text{cm}^3\)</td><td>\(50\,\text{cm}^3\)</td><td>\(75\,\text{cm}^3\)</td><td>\(100\,\text{cm}^3\)</td></tr> </table> c) \(48\) minutes.

- Look at where each graph crosses the y-axis. - How does the y-intercept relate to the constant term in each equation? - What does the slope represent in this pricing context? - Substitute a rental time into each equation to compare costs. - Look at where the graphs intersect and which graph is lowest before and after that point.

1. Match the graphs by their y-intercepts. Graph a passes through \((0, 0)\), so it represents \(f_1(x) = 4x\). Graph b has y-intercept \(6\), so it represents \(f_2(x) = 2x + 6\). Graph c has y-intercept \(9\), so it represents \(f_3(x) = x + 9\). 2. At \(x = 2\), \(f_1\) costs \(4 \cdot 2 = 8\), \(f_2\) costs \(2 \cdot 2 + 6 = 10\), and \(f_3\) costs \(2 + 9 = 11\). Plan \(f_1\) is least expensive at \(\$8\), and Plan \(f_3\) is most expensive at \(\$11\), so the savings are \(\$3\). 3. The three plans have the same cost at \(x = 3\), where each costs \(\$12\). For \(0 \le x < 3\), Plan \(f_1\) has the lowest cost. At \(x = 3\), all three plans tie. For \(x > 3\), Plan \(f_3\) has the lowest cost.

a) a: \(f_1(x) = 4x\); b: \(f_2(x) = 2x + 6\); c: \(f_3(x) = x + 9\) b) Plan \(f_1\) costs \(\$8\) and is least expensive. The savings compared with Plan \(f_3\) at \(\$11\) are \(\$3\). c) Plan \(f_1\) for \(0 \le x < 3\); all three plans at \(x = 3\); Plan \(f_3\) for \(x > 3\).

- Read the graph's y-intercept and use two graph points to find the rate of change of \(f\). - The verbal description of \(g\) states both its starting output and its change per input unit. - For part b, evaluate both functions at the same input before comparing their outputs.

1. From the graph, \(f\) has initial value \(12\). Using \((0,12)\) and \((6,0)\), its rate of change is \(\frac{0-12}{6-0}=-2\). 2. The verbal description gives \(g\) an initial value of \(9\) and a rate of change of \(-1.5\). 3. Thus \(f\) starts \(3\) units higher and decreases faster because \(-2<-1.5\). 4. At \(x=4\), the graph gives \(f(4)=4\). For \(g\), \(g(4)=9-1.5\cdot4=3\). Therefore, \(f(4)>g(4)\).

a) \(f\): initial value \(12\), rate \(-2\); \(g\): initial value \(9\), rate \(-1.5\). Function \(f\) starts higher and decreases faster. b) \(f\), because \(f(4)=4\) and \(g(4)=3\).

- One matching input-output pair does not determine an entire function. - Find the constant rate in the table and compare it with the coefficient of \(x\) in the equation. - Test a different input that appears in the table.

1. From the table, \(f\) increases by \(6\) when \(x\) increases by \(2\), so its rate of change is \(3\). 2. From the equation, \(g\) has rate of change \(2\). 3. The functions happen to have the same output at \(x=2\), but equal output at one input does not make two functions identical. Their different rates show that they cannot agree for every input. 4. For example, at \(x=0\), \(f(0)=2\) while \(g(0)=4\).

Jordan's conclusion is incorrect. Function \(f\) has rate \(3\), while \(g\) has rate \(2\). They agree at \(x=2\) only at that input; for example, \(f(0)=2\) and \(g(0)=4\).

- Read Nora's starting number and weekly gain from the dots. - Compare the input and output gaps in Eli's table. - For weeks after the gift, how should you adjust Nora's original count?

1. The dots show that Nora starts with \(5\) books and adds \(2\) each week. At week \(3\), she has \(5+2\cdot3=11\) before the gift and \(21\) after it. 2. Eli adds \((9-1)\div2=4\) books per week, starting with \(1\). His count at week \(w\) is \(1+4w\). 3. After week \(3\), Nora has \(15+2w\) books, including the gift. Eli has more when \(1+4w>15+2w\), or \(w>7\). 4. At the end of week \(7\), both have \(29\) books. Eli first has more at the end of week \(8\).

a) \(21\) books b) Week \(8\). Both have \(29\) books at week \(7\); Eli has more at week \(8\).