Appendix A — Answer Key, Chapter 7: Graphing and Evaluating Linear Functions

SOL A.F.1 (f, g, h) · Covers textbook Chapter 7 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 120 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below: a slope is written as a fraction before it is stepped, so an integer slope $m$ is $\tfrac{m}{1}$ and the run is $1$. An intercept is a point; a zero is a number. A bare-grid graph gets arrowheads on both ends; a context graph is a segment with marked endpoints. Viewing windows are not unique — any window that shows both intercepts with a readable scale is correct, and the windows given here are one reasonable choice each. A calculator's decimal is a rounding of an exact value, and the exact value is what gets written down. Where a question asks for both an algebraic result and a graphical confirmation, both are required for full credit, and a disagreement between them means neither is reported until the error is found.

The functions used repeatedly in the chapter, for reference:

• Figure 1 is $f(x) = \tfrac23 x - 4$, stepped from $(0,-4)$ to $(3,-2)$ to $(6,0)$
• Figure 2 is $h(x) = -2x + 3$, stepped right 1 down 2 to $(1,1)$ and left 1 up 2 to $(-1,5)$
• Figure 3 is $3x - 4y = 12$, with intercepts $(0,-3)$ and $(4,0)$
• Figure 4 is $f(x) = \tfrac12 x + 1$ at the even inputs $-4, -2, 0, 2, 4$
• Figure 5 is $y = -3$ beside $x = 4$
• Figure 6 is the bike rental, $C(h) = 5h + 20$ on $0 \le h \le 8$, at $1$ hour and $5$ dollars per cell
• Figure 7 is $f(x) = 25x - 140$, in the standard window and then in $x$ from $0$ to $10$ by $1$, $y$ from $-150$ to $100$ by $25$
• Figures 8 and 9 are both $f(x) = 2x - 3$ in the same window: figure 8 reads $f(4) = 5$, figure 9 recovers $x = -2$ from $f(x) = -7$
• Figure 10 is Plan A $10x + 60$ against Plan B $25x$, crossing at $(4,100)$
• Figure 11 is the pool, $W(t) = 400 - 25t$ on $0 \le t \le 16$, shown as context, equation, table, and graph
• Figure 12 is four blank grids, A through D, each running $-8$ to $8$ on both axes

Lesson 7.1 — Graphing by Hand from Slope-Intercept Form

Guided practice

1. First point $(0,-4)$, the $y$-intercept, read straight off the constant term. Slope $\tfrac23$. The steps produce $(3,-2)$ and then $(6,0)$.
2. "Right 3" is the run, the horizontal move, and it is the denominator of $\tfrac23$. "Up 2" is the rise, the vertical move, and it is the numerator. Together they say: from any point on the line, travel $3$ right and $2$ up and you are on the line again.
3. $f(6) = \tfrac23(6) - 4 = 4 - 4 = 0$. Because the output is $0$, the point $(6,0)$ is the $x$-intercept, and $6$ is the zero of the function.
4. $m = -2 = \dfrac{-2}{1}$. The run is $1$ and the rise is $-2$: right 1, down 2.
5. Left 1 and up 2 from $(0,3)$ lands on $(-1,5)$. Check: $h(-1) = -2(-1) + 3 = 2 + 3 = 5$ ✓
6. First point $(0,-2)$. The slope $3 = \tfrac31$ means right 1, up 3, giving $(1,1)$ and then $(2,4)$. Draw through all three with arrowheads on both ends.

Independent practice

7. a) $(0,1)$; right 1, up 4 (since $4 = \tfrac41$) b) $(0,5)$; right 1, down 1 (since $-1 = \tfrac{-1}{1}$) c) $(0,-2)$; right 5, up 3 d) $(0,4)$; right 3, down 2
8. Plot $(0,-3)$ and step right 2, up 1. Three integer points: $(0,-3)$, $(2,-2)$, $(4,-1)$. (Also acceptable: $(-2,-4)$, $(-4,-5)$, or any points satisfying $y = \tfrac12 x - 3$ with integer coordinates.)
9. Plot $(0,6)$ and step right 1, down 3. Three integer points: $(0,6)$, $(1,3)$, $(2,0)$. Note that $(2,0)$ is the $x$-intercept, so the graph hands you the zero.
10. The student inverted the fraction: $\tfrac25$ has run $5$ and rise $2$, not run $2$ and rise $5$. Going up $5$ and right $2$ graphs $y = \tfrac52 x + 1$ instead — a much steeper line through the same intercept. The correct step is right 5, up 2, landing on $(5,3)$.
11. From $(0,2)$, stepping right 4 and down 3 gives $(4,-1)$ and $(8,-4)$; stepping left 4 and up 3 gives $(-4,5)$. Four points: $(-4,5)$, $(0,2)$, $(4,-1)$, $(8,-4)$. Each checks in the rule — for instance $-\tfrac34(-4) + 2 = 3 + 2 = 5$ ✓
12. Slope measures the ratio of vertical change to horizontal change, and multiplying both parts of a ratio by $-1$ does not change it: $\dfrac{-2}{1} = \dfrac{2}{-1}$. So reversing both moves is still a step of the same slope, just travelled backwards. For $h(x) = -2x + 3$, right 1 and down 2 goes from $(0,3)$ to $(1,1)$, and left 1 and up 2 goes from $(0,3)$ to $(-1,5)$; substituting confirms both points lie on $h$.
13. First point $(0,500)$ — a full ream at the start of the shift. Slope $-25 = \tfrac{-25}{1}$, so the step is right 1 job, down 25 sheets. One sheet per cell is a bad choice because the vertical axis would need $500$ cells to reach the starting value; something like $25$ or $50$ sheets per cell fits the picture on a page and keeps every job's value on a gridline.
14. First point of $y = 2x$ is $(0,0)$; first point of $y = 2x + 3$ is $(0,3)$. Same: both have slope $2$, so both step right 1 up 2, and the two lines are exactly as steep as each other — they are parallel. Different: the $y$-intercept. Adding $3$ raises every output by $3$, which shifts the whole graph up $3$ without tilting it.
15. First point $(0,100)$ — the battery is full at $t = 0$. Slope $-8 = \tfrac{-8}{1}$: right 1 hour, down 8 percent. The graph reaches $0$ when $100 - 8t = 0$, so $8t = 100$ and $t = 12.5$ hours.
16. The student read the first number on the page instead of the constant term. In $y = mx + b$ the number that gives the first point is $b$, the constant, and $b$ is a $y$-coordinate. The correct first point is $(0,-4)$, and the step is right 1, up 3.

Exit ticket 7.1

17. First point $(0,5)$. Slope $-\tfrac43$: right 3, down 4. (Left 3, up 4 is the same step reversed and lands on $(-3,9)$.)
18. Plot $(0,-1)$ and step right 3, up 1: $(0,-1)$, $(3,0)$, $(6,1)$. The middle point is the $x$-intercept.
19. Two points determine exactly one line, so two are logically enough. A third is worth plotting because it is a free error check: if all three are collinear the arithmetic almost certainly worked, and if the third point misses the line through the first two, you know you have made a slip before you commit ink to a wrong graph.
20. $y$-intercept $(0,1)$. Slope $-2 = \tfrac{-2}{1}$: right 1, down 2, giving $(1,-1)$ and $(2,-3)$.

Lesson 7.2 — Intercepts, Tables, and the Two Special Lines

Guided practice

21. Let $x = 0$, which gives $-4y = 12$ and so $y = -3$; let $y = 0$, which gives $3x = 12$ and so $x = 4$. The two points are $(0,-3)$ and $(4,0)$.
22. Subtract $3x$: $-4y = -3x + 12$. Divide every term by $-4$: $y = \tfrac34 x - 3$. The slope $\tfrac34$ agrees with the plotted points, because travelling from $(0,-3)$ to $(4,0)$ is a run of $4$ and a rise of $3$. The $y$-intercept $(0,-3)$ matches the first plotted point.
23. The inputs $-4$, $-2$, $0$, $2$, $4$ — the even numbers. They were chosen because the slope is $\tfrac12$, and multiplying an even number by $\tfrac12$ gives a whole number, so every output lands exactly on a gridline.
24. The inputs $1$, $3$, $5$ produce $1.5$, $2.5$, $3.5$. Every point would sit halfway between two gridlines, which has to be estimated by eye — and a hand-drawn line through three estimated points is where a graphing error comes from.
25. For $y = -3$: find $-3$ on the $y$-axis and rule a horizontal line through it; every point on it has height $-3$. For $x = 4$: find $4$ on the $x$-axis and rule a vertical line through it; every point on it has first coordinate $4$. Neither needs a slope stepped, because neither has a slope to step — $y = -3$ has slope $0$ and $x = 4$ has no slope at all. $y = -3$ is the function; $x = 4$ fails the vertical line test.
26. One cell across is $1$ hour; one cell up is $5$ dollars. A $4$-hour rental costs $C(4) = 5(4) + 20 = 40$ dollars, and the figure shows that point sitting exactly on a grid corner $4$ cells right and $8$ cells up.

Independent practice

27. a) $x$-intercept $(5,0)$ from $2x = 10$; $y$-intercept $(0,2)$ from $5y = 10$ b) $x$-intercept $(2,0)$ from $3x = 6$; $y$-intercept $(0,-6)$ from $-y = 6$ c) $x$-intercept $(8,0)$ from $x = 8$; $y$-intercept $(0,-2)$ from $-4y = 8$ d) $x$-intercept $(-2,0)$ from $6x = -12$; $y$-intercept $(0,-4)$ from $3y = -12$
28. The denominator of $\tfrac23$ is $3$, so choose multiples of $3$.

Including $0$ is deliberate: it hands over the $y$-intercept $(0,2)$ with no work. Note that $(-3,0)$ is the $x$-intercept, so this table also gives the zero.

29. The denominator of $-\tfrac14$ is $4$, so choose multiples of $4$.

30. $y = -1$: horizontal through $-1$ on the $y$-axis; slope $0$; is a function. $x = 3$: vertical through $3$ on the $x$-axis; no slope (the run between any two of its points is $0$); not a function, because the single input $3$ is paired with every output at once.
31. $C(3) = 5(3) + 20 = 35$ dollars; $C(6) = 5(6) + 20 = 50$ dollars. Five dollars per cell is better than four because every cost the model produces — $20, 25, 30, 35, \dots$ — is a multiple of $5$, so every plotted point lands on a gridline. At $4$ dollars per cell, $C(1) = 25$ would sit a quarter of the way between two lines, and so would most of the other points.
32. Horizontal axis $0$ to $200$ miles at $20$ miles per cell ($10$ cells). Vertical axis $0$ to $110$ dollars at $10$ dollars per cell ($11$ cells). Endpoints $(0,45)$ — the fee before driving anywhere — and $(200, 105)$, since $D(200) = 0.30(200) + 45 = 60 + 45 = 105$. Draw a segment, because the domain is $0 \le m \le 200$.
33. Intercepts $(5,0)$ from $4x = 20$ and $(0,-4)$ from $-5y = 20$. The intercept method is faster. Converting first means dividing by $-5$ and carrying a fraction, $y = \tfrac45 x - 4$, and then stepping a run of $5$; the intercept method is two one-step substitutions and no fractions at all.
34. The student read $2x + 3y = 12$ as though it were slope-intercept form, but $2$ is the coefficient of $x$ in standard form and $3$ is the coefficient of $y$ — neither is a slope or an intercept. Correctly: let $x = 0$ to get $3y = 12$ and the point $(0,4)$; let $y = 0$ to get $2x = 12$ and the point $(6,0)$. (Converting gives $y = -\tfrac23 x + 4$, so the slope is even the wrong sign in the student's version.)
35. Let $y = 0$: $8x = 240$, so $x = 30$, giving $(30,0)$. Let $x = 0$: $12y = 240$, so $y = 20$, giving $(0,20)$. The point $(30,0)$ means selling $30$ wristbands and no shirts hits the $\$240$ goal; $(0,20)$ means selling $20$ shirts and no wristbands hits it too. Only the part of the line in the first quadrant is meaningful, because neither count can be negative.
36. Horizontal axis $0$ to $16$ minutes at $2$ minutes per cell ($8$ cells). Vertical axis $0$ to $400$ gallons at $50$ gallons per cell ($8$ cells). Both scales are chosen so the endpoints $(0,400)$ and $(16,0)$ land on corners, and so every four-minute reading — $300$, $200$, $100$ — does too.

Exit ticket 7.2

37. Let $y = 0$: $5x = 20$, so $x = 4$, giving $(4,0)$. Let $x = 0$: $2y = 20$, so $y = 10$, giving $(0,10)$.
38. The denominator of $\tfrac34$ is $4$, so use multiples of $4$: the inputs $-4$, $0$, $4$, $8$ give outputs $-4$, $-1$, $2$, $5$.
39. $y = 6$ is a function, with slope $0$. $x = -5$ is not a function and has no slope. Both are graphed by finding the constant on its own axis and ruling a straight line — $y = 6$ horizontally, $x = -5$ vertically.
40. Horizontal axis $0$ to $20$ minutes at $2$ minutes per cell. Vertical axis $0$ to $280$ gallons at $20$ gallons per cell. Endpoints $(0,30)$ and $(20,270)$, since $G(20) = 12(20) + 30 = 270$. Draw a segment: the story runs only from $0$ to $20$ minutes.

Lesson 7.3 — Graphing with Technology

Guided practice

41. The window reaches only from $-10$ to $10$ on each axis, but the $y$-intercept of the function is $-140$ and its zero is $5.6$. The intercept is $130$ units below the bottom of the screen, so the only part of the line inside the window is a short, nearly vertical stroke through one corner. A student reading that picture could not find either intercept, and might wrongly conclude the function has none.
42. $y$-intercept: the constant term, so $(0,-140)$. Zero: $25x - 140 = 0$ gives $25x = 140$ and $x = \tfrac{140}{25} = \tfrac{28}{5} = 5.6$. The intercept $-140$ fixed the bottom of the window at $-150$; the zero $5.6$ fixed the right-hand edge at $10$, comfortably past it.
43. One cell across is $1$ unit of $x$; one cell up is $25$ units of $y$.
44. $y$-intercept $(0,96)$; zero from $12x + 96 = 0$, so $x = -8$, giving $(-8,0)$. A window of $x$ from $-12$ to $4$ with a scale of $2$, and $y$ from $-20$ to $120$ with a scale of $20$, shows both. (Any window containing both points with a readable scale is correct.)
45. $f(5) = 25(5) - 140 = -15$ and $f(6) = 25(6) - 140 = 10$. The output changes from negative to positive between those two inputs, so the graph crosses the $x$-axis somewhere between $x = 5$ and $x = 6$ — which brackets the zero $5.6$ without solving anything.
46. A graph can verify that a point you computed really lies on the line, and that a solution you found sits where the picture says it should. It cannot establish an exact value, because a reading off a screen or a grid is only as precise as your eye and the calculator's rounding — the exact value has to come from the algebra.

Independent practice

47. a) Intercepts $(0,-45)$ and $(15,0)$, from $3x = 45$. Window: $x$ from $-5$ to $20$, scale $5$; $y$ from $-60$ to $20$, scale $10$. b) Intercepts $(0,8)$ and $(16,0)$, from $-\tfrac12 x + 8 = 0$ so $x = 16$. Window: $x$ from $-5$ to $20$, scale $5$; $y$ from $-5$ to $10$, scale $1$. c) Intercepts $(0,100)$ and $\left(-\tfrac12, 0\right)$, from $200x = -100$. Window: $x$ from $-2$ to $2$, scale $0.5$; $y$ from $-100$ to $500$, scale $100$. This one is steep, so the $x$-window must be narrow while the $y$-window is wide. d) Intercepts $(0,-2)$ and $(2,0)$. The standard window is fine here: $x$ and $y$ both from $-10$ to $10$, scale $1$. Not every function needs a custom window, and saying so is part of the answer.
48. $T(0) = 400$ and $T(52) = 35(52) + 400 = 1820 + 400 = 2220$. Window: $x$ from $0$ to $52$, scale $4$; $y$ from $0$ to $2400$, scale $200$. The zero of $T$ is negative, so it is outside the story and does not need to be on screen. $T(52) = \$2220$.
49. The most likely cause is a typing error: the function entered was $-2x + 5$ rather than $2x + 5$, because a positive slope of $2$ must rise from left to right. (A window cannot reverse a line's direction, so the window is not the suspect here.) To check, look at the entry line and re-read it against the printed problem, then confirm with the table feature: $f(1)$ should be $7$, not $3$.
50. $4x - 7 = 13$ gives $4x = 20$, so $x = 5$. To confirm, graph $f(x) = 4x - 7$ and check that the point $(5,13)$ is on it — either by tracing to $x = 5$ and reading $13$, or by using the table feature, which should print $13$ in the output column beside the input $5$.
51. The zero is real but off screen: $0.1x - 4 = 0$ gives $0.1x = 4$, so $x = 40$, and the standard window stops at $10$. The student mistook "not visible in this window" for "does not exist." Widening the window to $x$ from $-10$ to $50$ shows the crossing at $(40,0)$.
52. The table of $f(x) = -6x + 15$:

The output changes sign between $x = 2$ and $x = 3$, so the zero lies between them. Exactly: $-6x + 15 = 0$ gives $6x = 15$ and $x = \tfrac{15}{6} = \tfrac52 = 2.5$.

53. They almost certainly chose different viewing windows. The same line can look nearly vertical in one window and nearly horizontal in another, because changing the scale of one axis changes the apparent steepness. They should compare their window settings first — the four boundary numbers and the two scales — and only then suspect a typing error in the function itself.
54. $1.2m + 30 = 90$ gives $1.2m = 60$, so $m = 50$ miles. The graphical check: graph $C(m) = 1.2m + 30$ in a window reaching $m = 60$ and $C = 100$, and confirm that the point $(50, 90)$ lies on the line — by tracing to $50$ and reading $90$, or by reading the table at $m = 50$.
55. A window is only useful if the features you want are inside it, and you cannot know where those features are until you have located them algebraically. The two numbers to compute first are the $y$-intercept (the constant term, which fixes how far up or down the window must reach) and the zero (the solution of $mx + b = 0$, which fixes how far left or right it must reach).
56. Exact zero: $3x - 10 = 0$ gives $3x = 10$, so $x = \tfrac{10}{3}$. The screen shows a decimal approximation of that fraction, cut off at the width of the display; $3.3333333$ is not a different number from $\tfrac{10}{3}$, it is a rounded picture of it. Write $\tfrac{10}{3}$, and use the decimal only as a confirmation that the algebra and the screen agree.

Exit ticket 7.3

57. $y$-intercept $(0,120)$; zero from $-15x + 120 = 0$, so $x = 8$, giving $(8,0)$. Window: $x$ from $-2$ to $12$, scale $2$; $y$ from $-40$ to $140$, scale $20$.
58. $f(7) = -15(7) + 120 = -105 + 120 = 15$. Graphical check: in the window above, trace to $x = 7$ and confirm the screen reports $15$, or read the table at $x = 7$. Note that $(7,15)$ sits just left of the zero at $x = 8$, which is a second sanity check — the output should still be positive there.
59. The rule: if the algebra and the graph disagree, one of them is wrong and you are not finished. Do not average them and do not choose the one you prefer. Check first whether the function was typed correctly — a missing parenthesis is the most common cause — and second whether the window is hiding the feature you are looking at. Then re-solve the algebra.
60. The student copied a rounded decimal instead of recognizing the exact value behind it. $5x - 15 = 0$ gives $5x = 15$, so the zero is exactly $3$. The display $2.9999999$ is the calculator's rounding of $3$; a zero that comes out to a whole number in the algebra is a whole number, and that is what belongs in the answer.

Lesson 7.4 — Determining $f(x)$, and Recovering $x$

Guided practice

61. $f(4) = 5$. The dashed path travels up from $4$ on the $x$-axis until it meets the line, and then across from that meeting point to the $y$-axis, where it arrives at $5$. The meeting point is $(4,5)$.
62. $f(4) = 2(4) - 3 = 8 - 3 = 5$ ✓ — the same value the graph gave.
63. $f(-1) = 2(-1) - 3 = -2 - 3 = -5$
64. The input $-2$ produced it. The path travels across from $-7$ on the $y$-axis until it meets the line, and then down the vertical to the $x$-axis, arriving at $-2$. The meeting point is $(-2,-7)$.
65. $2x - 3 = -7$, so $2x = -4$ and $x = -2$ ✓
66. Item 61 hands you an input and asks for the output, so you substitute. Item 64 hands you an output and asks for the input, so you solve an equation. What they share is the point: both answers are coordinates of a single point on the graph, and each question gives you one coordinate and asks for the other.

Independent practice

67. a) $f(0) = -3(0) + 8 = 8$ b) $f(2) = -3(2) + 8 = -6 + 8 = 2$ c) $f(-4) = -3(-4) + 8 = 12 + 8 = 20$ — the parentheses are what keep the double negative straight d) $f\!\left(\tfrac13\right) = -3\!\left(\tfrac13\right) + 8 = -1 + 8 = 7$
68. a) $f(10) = \tfrac12(10) - 6 = 5 - 6 = -1$ b) $f(-8) = \tfrac12(-8) - 6 = -4 - 6 = -10$ c) $f(0) = -6$ d) $f(3) = \tfrac32 - 6 = -\tfrac92$, which is $-4.5$ — an output need not be a whole number
69. a) $-3x + 8 = -7$, so $-3x = -15$ and $x = 5$ b) $-3x + 8 = 8$, so $-3x = 0$ and $x = 0$ c) $-3x + 8 = 23$, so $-3x = 15$ and $x = -5$ d) $-3x + 8 = 0$, so $-3x = -8$ and $x = \tfrac83$ — this one is the zero of the function
70. a) $4x + 1 = 13$, so $4x = 12$ and $x = 3$ b) $4x + 1 = -11$, so $4x = -12$ and $x = -3$ c) $4x + 1 = 1$, so $4x = 0$ and $x = 0$ d) $4x + 1 = 3$, so $4x = 2$ and $x = \tfrac12$ — a recovered input need not be a whole number either
71. Reading the graph: $f(0) = -3$, $f(1) = -1$, $f(2) = 1$. Confirming: $2(0) - 3 = -3$ ✓, $2(1) - 3 = -1$ ✓, $2(2) - 3 = 1$ ✓. Notice $f(0)$ is the $y$-intercept, so the graph gave it away before any arithmetic.
72. $f(x) = 1$ at $x = 2$, since $2x - 3 = 1$ gives $2x = 4$. And $f(x) = -3$ at $x = 0$, since $2x - 3 = -3$ gives $2x = 0$. Both start from the $y$-axis, because in each case the number given is an output.
73. $C(12) = 2.5(12) + 3.5 = 30 + 3.5 = 33.5$. A $12$-mile ride costs $\$33.50$. For the second: $2.5m + 3.5 = 28.5$ gives $2.5m = 25$, so $m = 10$. A fare of $\$28.50$ was paid for a $10$-mile ride. The units differ because the two coordinates measure different things — the output is in dollars, the input in miles.
74. $T(6) = 68 - 4(6) = 68 - 24 = 44$ degrees after $6$ hours. And $68 - 4h = 20$ gives $4h = 48$, so $h = 12$ hours. The room reaches $20$ degrees after $12$ hours.
75. The student substituted when the question called for solving. The number $10$ was given as an output — the problem wrote $f(x) = 10$, with the $10$ outside the parentheses — so it belongs on the right-hand side of an equation. Computing $f(10)$ answers the different question "what is the output when the input is $10$?" Correctly: $2x + 4 = 10$ gives $2x = 6$, so $x = 3$.
76. Finding $f(x)$ from a given $x$ is pure substitution: every letter in the rule gets replaced by a number, and what remains is arithmetic with no unknown in it. Recovering $x$ leaves the unknown in place — the rule still contains $x$, and the given output tells you only what the rule must equal. A statement of the form "expression containing $x$ equals a number" is precisely an equation, so it has to be solved.
77. $f(8) = 0.75(8) - 2.5 = 6 - 2.5 = 3.5$. The table feature confirms it directly: set the table to start at $x = 8$ (or step by $1$ from $0$ and read down to the row $x = 8$) and check that the output column shows $3.5$. If it shows anything else, the function was typed wrong — most likely as $0.75x - 25$ or $0.75(x - 2.5)$.
78. $C(5) = 5(5) + 20 = 45$ dollars — on the graph, up from $5$ on the horizontal axis to the segment, then across to $45$ on the vertical axis. For $\$55$: across from $55$ to the segment, then down to $7$ on the horizontal axis; algebraically $5h + 20 = 55$ gives $5h = 35$, so $h = 7$ hours. Both readings land on grid corners, because the scale was chosen so they would.

Exit ticket 7.4

79. $f(-3) = -2(-3) + 9 = 6 + 9 = 15$. And $-2x + 9 = 1$ gives $-2x = -8$, so $x = 4$.
80. $f(4) = 5$, and $f(x) = 5$ at $x = 4$. One marked point answers both because the point is $(4,5)$: the pair of numbers $4$ and $5$ is a single fact about the function, and the two questions differ only in which of the two numbers you were given. Reading the point left-to-right answers the first; reading it right-to-left answers the second.
81. $B(9) = 250 - 18(9) = 250 - 162 = 88$ dollars after $9$ weeks. And $250 - 18w = 34$ gives $18w = 216$, so $w = 12$: the balance reaches $\$34$ at week $12$.
82. One path starts at the $x$-axis, goes vertically to the line, and ends at the $y$-axis — that is the path for finding $f(x)$ from a given input. The other starts at the $y$-axis, goes horizontally to the line, and ends at the $x$-axis — that is the path for recovering $x$ from a given output. Both paths use the same corner point on the line; only the order of the two travels is reversed.

Lesson 7.5 — Comparing the Four Representations

Guided practice

83. $W(t) = 400 - 25t$.

84. The table makes the constant rate of change visible directly: every step of $4$ minutes drops $W$ by exactly $100$ gallons, four times in a row. The equation contains that fact in the coefficient $-25$, but it does not display it as a repeated pattern — you have to know to look at the coefficient and know what it means. Equal differences over equal steps is what linearity looks like, and a table is where you see it.
85. The trend, immediately: one glance says the water is running out, steadily, and will be gone. No reading, no arithmetic. The graph also shows the domain and range at a glance, as the horizontal and vertical extent of the segment.
86. $W(10) = 400 - 25(10) = 400 - 250 = 150$ gallons. The equation was the right choice, because it is the only representation that gives an exact value at an arbitrary input. The table does not contain $t = 10$ at all, and reading $t = 10$ off the graph gives an estimate near $150$ but cannot certify it — the point falls between the plotted gridlines of the vertical scale.
87. At $2$ hours: Plan B costs $25(2) = \$50$ and Plan A costs $10(2) + 60 = \$80$, so B is cheaper. At $6$ hours: Plan A costs $10(6) + 60 = \$120$ and Plan B costs $25(6) = \$150$, so A is cheaper. The reversal happens because A has the higher starting cost but the lower rate.
88. Set the two outputs equal: $10x + 60 = 25x$. Subtracting $10x$ gives $60 = 15x$, so $x = 4$, and then $25(4) = 100$. The crossing is $(4,100)$, exactly where the figure marks it — one equation in one variable, solved the way Chapter 2 taught, confirms what the picture showed.

Independent practice

89. Algebraic: $f(x) = 6x + 15$, where $x$ is hours. Tabular:

**Graphical:** a segment starting at $(0,15)$ and rising $6$ dollars for each hour; at $1$ hour and $5$ dollars per cell it climbs a little over one cell per hour, and the domain is restricted to non-negative hours. **Contextual:** "The service charges $\$6$ per hour of work, on top of a $\$15$ trip fee that is paid even for a zero-hour visit." The slope carries the units **dollars per hour**, which only the sentence supplies.

90. Rates of change: $f$ has slope $4$; $g$ rises $8$ for every $2$ across, so $\tfrac{8}{2} = 4$ as well — equal. Initial values: $f$ has $(0,20)$; $g$'s table contains the input $0$ with output $30$, so $(0,30)$ — $g$ starts higher. Because the slopes are equal and the intercepts are not, the graphs are parallel and never meet; $g$ is greater than $f$ by exactly $10$ at every input, forever.
91. $f$ has slope $-5$ and initial value $40$. For $g$, the slope is $\dfrac{0 - 25}{5 - 0} = -5$ and the initial value is $25$. The slopes are equal and $f$ starts higher. So the two graphs are parallel lines both falling at the same rate; $f$ stays $15$ above $g$ at every input, and they never cross.
92. Ranking, greatest rate first: the table (rate $4$), then the equation $y = 3x - 1$ (rate $3$), then the context (rate $2.50$ dollars per item). What made each visible: in the equation, the coefficient of $x$; in the table, the constant difference in the output over a step of $1$ in the input; in the context, the phrase "per item," which is what a rate sounds like in words — and which is the only one of the three that comes with units.
93. Gym A: $A(c) = 5c + 25$. Gym B, from the table, rises $20$ for every $2$ classes, so its slope is $10$, and its row at $c = 0$ gives an initial value of $10$: $B(c) = 10c + 10$. At $3$ classes, $A(3) = 40$ and $B(3) = 40$ — the same $\$40$. Setting $5c + 25 = 10c + 10$ gives $15 = 5c$, so $c = 3$, confirming it. For someone taking $6$ classes a month, Gym A is better: $A(6) = 55$ against $B(6) = 70$. A has the lower rate per class, so past the crossing its higher monthly fee is more than paid back.
94. For a zero exactly, use the algebraic representation: set $mx + b = 0$ and solve, which produces an exact value even when it is a fraction like $\tfrac83$. To see a trend instantly, use the graphical representation: a rising or falling line communicates direction and steepness before you have read a single number, which no equation or table does.
95. The student read the first row of the table as though it were the $y$-intercept, but the table does not contain the input $0$ — it starts at $x = 1$. The slope is $3$, since the output rises $3$ for each step of $1$. Stepping back one row from $(1,7)$ gives $(0,4)$, so the $y$-intercept is $(0,4)$ and the equation is $y = 3x + 4$. Checking: $3(2) + 4 = 10$ ✓ and $3(3) + 4 = 13$ ✓
96. Equation: $C(t) = 0.25t + 4$ dollars for $t$ minutes. Table:

**Graph:** a segment starting at $(0,4)$ and rising; $10$ minutes per cell across and $1$ dollar per cell up puts $(0,4)$, $(20,9)$, and every other ten-minute reading on or near a corner, and the domain begins at $t = 0$ because a ride cannot last a negative number of minutes. **Slope:** $\$0.25$ per minute — every extra minute of riding adds a quarter to the bill.

97. Plan A has the larger $y$-intercept, $(0,60)$. That $\$60$ is the membership fee, paid before any court time at all — which is why Plan A costs money even at zero hours. Plan A still wins in the long run because the $y$-intercept is a one-time amount while the slope is charged per hour: A's rate of $\$10$ per hour is $\$15$ per hour less than B's, so after $4$ hours the accumulated saving on the rate has covered the $\$60$ fee, and every hour after that widens A's advantage.
98. Conclude that one of the two representations is wrong, because a single function cannot have two different outputs at the same input — that would violate the definition of a function from Chapter 4. Do not average the two values and do not assume the equation wins by default. Recompute $f(3) = 3(3) + 1 = 10$ carefully, then check a second table entry against the equation: if the rest of the table agrees with the rule, the $11$ is a copying error in the table; if the whole table is shifted, the equation was written down wrong.

Exit ticket 7.5

99. The table of $f(x) = -4x + 18$:

Slope $-4$ (each step of $1$ drops the output by $4$); $y$-intercept $(0,18)$, visible in the table's first column; zero from $-4x + 18 = 0$, so $4x = 18$ and $x = \tfrac92 = 4.5$.

100. $y = 7x + 2$ has slope $7$ and initial value $2$. The table rises $5$ for each step of $1$, so its slope is $5$, and its row at $x = 0$ gives an initial value of $9$: the function is $y = 5x + 9$. Greater rate: $y = 7x + 2$. Greater start: the table's function. They agree where $7x + 2 = 5x + 9$, so $2x = 7$ and $x = \tfrac72 = 3.5$; both then output $7(3.5) + 2 = 26.5$.
101. For the exact cost of $17$ hours, hand over the equation — substituting gives an exact number, while a graph gives an estimate and a table that stops short of $17$ gives nothing. For when two plans cross, hand over the graph — the crossing is a single visible point, and the picture also shows which plan is cheaper on each side of it, which the equations do not display.
102. Algebraic: exact values at any input, and the slope and intercept read straight off the symbols. Graphical: the trend, and the comparison between two functions, seen at a glance. Tabular: the constant difference over equal steps, which is what makes the function linear. Contextual: the meaning and the units, which no equation, graph, or table supplies on its own.

Chapter 7 Review

Part A — Graphing without technology

103. First point $(0,-4)$. Slope $\tfrac23$: right 3, up 2. Two more points: $(3,-2)$ and $(6,0)$. Draw through all three with arrowheads. The third point is also the $x$-intercept, so this graph reports the zero $6$ for free.
104. Let $y = 0$: $4x = 12$, so $x = 3$, giving $(3,0)$. Let $x = 0$: $-3y = 12$, so $y = -4$, giving $(0,-4)$. Draw the line through the two points.
105. The denominator of $-\tfrac25$ is $5$, so use multiples of $5$.

 Including $0$ gives the $y$-intercept $(0,3)$ at no cost.

106. $y = -4$: horizontal, slope $0$, is a function. $x = 1$: vertical, no slope, not a function. Graph each by finding the constant on its own axis and ruling a straight line through it.
107. Horizontal axis $0$ to $10$ days at $1$ day per cell. Vertical axis $0$ to $750$ miles at $50$ miles per cell ($15$ cells). Endpoints $(0,120)$ — the reading before the route starts — and $(10,720)$, since $S(10) = 60(10) + 120 = 720$. Draw a segment, because the route lasts exactly $10$ days. At $50$ miles per cell, $120$ falls between gridlines; a scale of $60$ miles per cell would put every daily reading on a corner and is the better choice if you want the plotted points exact.