Appendix A — Answer Key, Chapter 5: Linear Functions: Characteristics and Forms

SOL A.F.1 (a, b, c) · Covers textbook Chapter 5 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 124 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below: an intercept is a point and is written as an ordered pair; a zero is a number, and it is the $x$-coordinate of the $x$-intercept. A linear function is $f(x) = mx + b$ with slope $m$ and $y$-intercept $(0,b)$. Domains and ranges of continuous graphs are given in words or with inequalities; interval notation is never required. A vertical line $x = a$ is not a function and has no slope.

The functions used repeatedly in the chapter, for reference:

• Figure 1 is $f(x) = 2x - 4$, with intercepts $(0,-4)$ and $(2,0)$
• Figure 2 is the horizontal line $y = 3$ beside the vertical line $x = -2$
• Figure 3 is the cooler, $V(x) = -2x + 8$ on the domain $0 \le x \le 4$
• Figure 4 is $f(x) = \tfrac34 x - 3$, with intercepts $(0,-3)$ and $(4,0)$ and a rise-$3$, run-$4$ triangle
• Figure 5 is $f(x) = -2x + 6$, with zero $3$
• Figure 6 is the parent, $y = x$; Figure 7 shows $y = 3x$, $y = \tfrac13 x$, $y = -x$, $y = -2x$, $y = x+3$, and $y = x-4$ against it
• Figure 8 is $y = 2x - 4$, also written $2x - y = 4$ and $y - 2 = 2(x-3)$, through $(3,2)$
• Figure 9 is $3x + 4y = 12$; Figure 10 is $x = 3$ beside $y = -2$

Lesson 5.1 — Domain and Range of a Linear Function

Guided practice

1. Domain: all real numbers. Range: all real numbers. The slope $2$ is not zero, so the graph is an unbroken slanted line with no ends: every real number can be substituted, and the climbing line eventually reaches every height.
2. The arrowheads say the line continues past the edge of the grid rather than stopping there. Without them the drawn portion would be a segment running only from about $x = -1$ to $x = 5$, and its domain and range would both be restricted to that stretch.
3. Domain: all real numbers. Range: the single value $3$. Every input works, but the line never leaves the height $3$.
4. No, $x = -2$ is not a function. The single input $-2$ is paired with every output at once, so the question "what is the output at $-2$?" has infinitely many answers — the vertical line test fails against the line itself. It has no slope because the run between any two of its points is $0$, and a rise cannot be divided by $0$.
5. Domain $0 \le x \le 4$ minutes; range $0 \le y \le 8$ liters.
6. The equation $-2x + 8$ accepts any input: it would report $-32$ liters at $x = 20$ and $14$ liters at $x = -3$. Neither is a fact about the cooler, because pouring cannot begin before it begins and a cooler cannot hold a negative amount of water. The situation, not the equation, is what stops the domain at $0$ and $4$.

Independent practice

7. a) Domain all real numbers; range all real numbers ($m = -5 \ne 0$) b) Domain all real numbers; range all real numbers ($m = \tfrac14 \ne 0$) c) Domain all real numbers; range the single value $7$ — this is a horizontal line, $m = 0$ d) Domain all real numbers; range all real numbers ($m = 1$)
8. Domain $0 \le t \le 8$ hours; range $0 \le H \le 12$ inches. The left endpoint is fixed by the moment the candle is lit, when no time has passed and the candle is its full $12$ inches. The right endpoint is fixed by the candle burning out: $12 - 1.5t = 0$ gives $1.5t = 12$, so $t = 8$ hours, and there is no candle to model after that.
9. Domain $\{1,2,3,4,5,6\}$ passengers; range $\{2.50, 5, 7.50, 10, 12.50, 15\}$ dollars. The graph is six separate dots because a van cannot carry $2.5$ passengers — the situation admits only whole numbers of people, so nothing is being claimed between the plotted inputs.
10. A nonzero slope means the line is slanted, and a slanted line drawn without ends keeps climbing (or falling) forever. Name any output $y$ you like; solving $mx + b = y$ for $x$ is possible for every $y$ precisely because $m \ne 0$, so some input produces it. The values of $m$ and $b$ change where the line sits and how steeply it climbs, but not the fact that it eventually passes every height.
11. Domain $0 \le x \le 4$: the pouring can last anywhere from $0$ minutes up to $4$ minutes, and no longer, because after $4$ minutes there is nothing left to pour. Range $0 \le y \le 8$: the amount of water in the cooler is somewhere between empty and its full $8$ liters at every moment of the story.
12. The student confused where the graph crosses an axis with where the graph stops. The point $(0,-4)$ is only the crossing of the $y$-axis; the line continues below it forever, producing outputs far less than $-4$ — at $x = -10$ the output is $-24$. Correct range: all real numbers.

Exit ticket 5.1

13. Domain: all real numbers. Range: all real numbers. The slope $-3$ is not zero.
14. Domain: all real numbers. Range: the single value $-4$. It is a horizontal line, slope $0$.
15. No. Every point of $x = 5$ has the input $5$, so that one input is paired with infinitely many outputs. It has no slope, because the run between any two of its points is $0$.
16. Arrowheads say the graph never stops, so the domain is all real numbers regardless of how much of the line the grid shows. Two filled-in endpoints say the graph does stop, at inputs that are themselves included, so the domain is the stretch of real numbers between those two $x$-values, endpoints included. The drawn ink can look the same on the page; the arrowheads and the endpoints are what tell you which claim is being made.

Lesson 5.2 — Slope and the Two Intercepts

Guided practice

17. Slope $m = \tfrac34$; $y$-intercept $(0,-3)$; $x$-intercept $(4,0)$.
18. They are the two legs of the dashed triangle drawn from $(0,-3)$ across to $(4,-3)$ and up to $(4,0)$. The run $4$ is how far right you travel and the rise $3$ is how far up you then have to go to get back on the line, so $m = \dfrac{\text{rise}}{\text{run}} = \dfrac{3}{4}$.
19. The graph crosses the $y$-axis at $(0,-3)$, so $b = -3$.
20. $m = \dfrac{10 - 2}{3 - (-1)} = \dfrac{8}{4} = 2$
21. $m = \dfrac{1 - 7}{6 - 2} = \dfrac{-6}{4} = -\dfrac{3}{2}$
22. Slope $-5$; $y$-intercept $(0,8)$.

Independent practice

23. a) $m = 4$; $(0,-9)$ b) $m = -\tfrac23$; $(0,5)$ c) $m = 1$; $(0,0)$ — this is the parent function, $y = 1x + 0$ d) $m = 0$; $(0,-7)$ — a horizontal line
24. a) $\dfrac{11-3}{4-0} = \dfrac{8}{4} = 2$ b) $\dfrac{-4-5}{1-(-2)} = \dfrac{-9}{3} = -3$ c) $\dfrac{-1-(-1)}{2-(-6)} = \dfrac{0}{8} = 0$ — a horizontal line d) $\dfrac{9-4}{3-3} = \dfrac{5}{0}$, undefined — the two points share the input $3$, so this is the vertical line $x = 3$, which has no slope and is not a function
25. $y$-intercept $(0,-10)$, read from the equation. For the $x$-intercept, $5x - 10 = 0$ gives $5x = 10$ and $x = 2$, so $(2,0)$.
26. $y$-intercept $(0,3)$. For the $x$-intercept, $-\tfrac14 x + 3 = 0$ gives $\tfrac14 x = 3$ and $x = 12$, so $(12,0)$.
27. Slope $4$: each time $x$ increases by $1$, $y$ increases by $4$, and the $x$-steps are equal, so that constant $y$-step is the rate of change. $y$-intercept $(0,-5)$: the table already contains the input $0$, and its output is $-5$, so the point is read straight off the first column. (Any two columns give the same slope: from $(1,-1)$ to $(3,7)$, $\tfrac{7-(-1)}{3-1} = \tfrac{8}{2} = 4$.)
28. The slope $0.05$ is a rate of $\$0.05$ per minute: every additional minute of calling adds five cents to the monthly bill. The $y$-intercept $(0,20)$ is the starting value: a month with zero minutes of calling still costs $\$20$, which is the monthly fee.
29. The slope $-15$ means the tank loses $15$ gallons every minute; it is negative because the amount of water is going down. The $y$-intercept $(0,240)$ means the tank held $240$ gallons at the moment draining began. For the $x$-intercept, $240 - 15t = 0$ gives $15t = 240$ and $t = 16$, so $(16,0)$: the tank is empty after $16$ minutes.
30. The student read the two numbers in the order they appear on the page instead of reading which one multiplies $x$. The slope is the coefficient of $x$ and its sign travels with it, and the $y$-intercept is the constant term. Rewriting the equation as $y = -2x + 3$ makes both visible. Correct slope $-2$; correct $y$-intercept $(0,3)$.

Exit ticket 5.2

31. Slope $-\tfrac12$; $y$-intercept $(0,6)$.
32. $m = \dfrac{4 - 8}{5 - (-3)} = \dfrac{-4}{8} = -\dfrac12$
33. $y$-intercept $(0,12)$. For the $x$-intercept, $3x + 12 = 0$ gives $3x = -12$ and $x = -4$, so $(-4,0)$.
34. The slope $8$ is a rate of $\$8$ per ticket: each additional ticket added to the order costs $\$8$ more. The $y$-intercept $(0,45)$ is a fixed $\$45$ charge on the order itself — a service or processing fee that is paid regardless of how many tickets are bought.

Lesson 5.3 — The Zero of a Linear Function

Guided practice

35. The zero is $3$. The point that names it is the $x$-intercept $(3,0)$.
36. Both sentences say that the input $3$ produces the output $0$: $f(3) = 0$ says it in function language, and "the graph passes through $(3,0)$" says it in coordinate language, since a point of the graph is an input paired with its output.
37. $4x - 20 = 0$, so $4x = 20$ and $x = 5$. The zero is $5$; the $x$-intercept is $(5,0)$.
38. $-3x + 7 = 0$, so $-3x = -7$ and $x = \tfrac73$. The zero is $\tfrac73$; the $x$-intercept is $\left(\tfrac73, 0\right)$.
39. The zero is $4$. The figure shows it as the marked point $(4,0)$ where the line crosses the $x$-axis, and the label beside it says so directly. Checking algebraically: $\tfrac34 x - 3 = 0$ gives $\tfrac34 x = 3$, so $x = 4$.
40. No. The line $y = 3$ has slope $0$ and sits three units above the $x$-axis at every input, so it never touches the $x$-axis and no input ever produces the output $0$.

Independent practice

41. a) $x - 9 = 0$, so the zero is $9$; $x$-intercept $(9,0)$ b) $-5x = 0$, so the zero is $0$; $x$-intercept $(0,0)$, which is also the $y$-intercept, since the line passes through the origin c) $2x + 7 = 0$, so $2x = -7$ and the zero is $-\tfrac72$; $x$-intercept $\left(-\tfrac72, 0\right)$ d) $\tfrac13 x - 4 = 0$, so $\tfrac13 x = 4$ and the zero is $12$; $x$-intercept $(12,0)$
42. $320 - 40w = 0$ gives $40w = 320$, so the zero is $8$. It means the savings run out after $8$ weeks — that is the week in which the balance reaches $\$0$.
43. $-2x + 8 = 0$ gives $2x = 8$, so the zero is $4$. It means the cooler is empty after $4$ minutes of pouring. The domain of the model ends at exactly that input because after the cooler is empty there is nothing left to describe: the equation would keep reporting negative liters, which is not a fact about any cooler. In a draining model the zero and the right-hand endpoint of the realistic domain are the same number.
44. The student reported the constant term, which gives the $y$-intercept $(0,-10)$ — the output at $x = 0$, not the input that makes the output $0$. Solving $2x - 10 = 0$ gives $2x = 10$, so the correct zero is $5$ and the correct $x$-intercept is $(5,0)$.

Exit ticket 5.3

45. $6x + 18 = 0$, so $6x = -18$ and the zero is $-3$.
46. $-\tfrac14 x + 5 = 0$, so $\tfrac14 x = 5$ and the zero is $20$.
47. $90 - 12t = 0$ gives $12t = 90$, so the zero is $t = 7.5$. It means the drone reaches the ground $7.5$ seconds after it begins descending.
48. All three ask for the input whose output is $0$. The zero names that input as a number, the $x$-intercept names the point $(x, 0)$ where the graph sits at height zero, and solving $mx + b = 0$ is the algebra that produces it. One fact, described in function language, in coordinate language, and in equation language.

Lesson 5.4 — The Parent Function $y = x$

Guided practice

49. Slope $m = 1$; $y$-intercept $(0,0)$.
50. $(-3,-3)$, $(-1,-1)$, $(0,0)$, $(1,1)$, and $(3,3)$. In every one, the two coordinates are the same number.
51. It shows a rise of $1$ over a run of $1$: travel one unit right from a point of the line and you must go one unit up to land back on it. So $m = \tfrac11 = 1$.
52. Domain: all real numbers. Range: all real numbers.
53. The zero is $0$. It is carried by the point $(0,0)$, which is simultaneously the $x$-intercept and the $y$-intercept, because the line passes through the origin.
54. A parent function is the simplest member of a family — the one every other member can be described as a modification of. $y = x$ is the parent of the linear family because writing the general linear function as $y = mx + b$ shows there are only two things to change, $m$ and $b$, and $y = x$ is the case $m = 1$, $b = 0$. Every other linear function is reached from it by choosing a different slope, a different $y$-intercept, or both.

Independent practice

55. $x$	$-4$	$-2$	$0$	$5$	$7$
    $y$	$-4$	$-2$	$0$	$5$	$7$

    Each output is a copy of its input.

56. $(-6,-6)$ is on the graph, because the rule requires the output to equal the input and $-6 = -6$. $(2,-2)$ is not, because the input $2$ must produce the output $2$, and $-2 \ne 2$. (The point $(2,-2)$ lies on $y = -x$ instead.)

57. $m = 1$ and $b = 0$. Written in full, $y = x$ is $y = 1x + 0$.

58. The slope $1$ is a rate of one credit hour per volunteer hour: each hour a student volunteers earns exactly one hour of credit, no more and no less. The $y$-intercept $(0,0)$ means a student who volunteers nothing has no credit — nobody is given a head start, and the tally begins at zero.

Exit ticket 5.4

59. Slope $1$; $y$-intercept $(0,0)$; zero $0$; domain all real numbers; range all real numbers.
60. Yes. The rule of the parent function is that each output equals its own input, and $-9 = -9$.
61. The line $y = 0$ is not a parent but a special case: it is the horizontal line along the $x$-axis, with slope $0$, and no choice of $b$ alone will ever give it a nonzero slope. A parent has to be the member from which every other member is reachable by the family's transformations, and the linear family's dials are the slope $m$ and the intercept $b$. The correct parent is $y = x$, the case $m = 1$, $b = 0$.
62. $y = x$ is the parent because every linear function is $y = mx + b$, and $y = x$ is that expression with $m = 1$ and $b = 0$ — the plainest choice of both dials, so every other linear function is a transformation of it. One transformation of it is $y = 3x$, the parent with its slope tripled. (Any linear function other than $y = x$ itself is an acceptable answer.)

Lesson 5.5 — Transformations of the Parent Function

Guided practice

63. The equations are $y = 3x$, with slope $3$, and $y = \tfrac13 x$, with slope $\tfrac13$. Both share the $y$-intercept $(0,0)$ with the parent.
64. $y = 3x$ is steeper and $y = \tfrac13 x$ is less steep. Steepness is the size of the slope: $|3| = 3 > 1$, so that line climbs three units per unit across instead of the parent's one, while $|\tfrac13| < 1$, so that line climbs only a third of a unit per unit across and lies flatter.
65. Both lines are reflections of the parent across the $x$-axis — they fall from left to right where the parent rises. $y = -x$ has slope $-1$ and $y = -2x$ has slope $-2$, so the second is reflected and steeper, since $|-2| = 2 > 1$.
66. The $y$-intercept changed: $y = x + 3$ crosses at $(0,3)$ and $y = x - 4$ crosses at $(0,-4)$, instead of the parent's $(0,0)$. The slope stayed the same at $1$ — both lines are exactly as steep as the parent, just sitting higher or lower.
67. The slope is multiplied by $5$, from $1$ to $5$, so the line is steeper than the parent. The $y$-intercept is unchanged at $(0,0)$, because nothing was added to the rule and $5 \cdot 0 = 0$.
68. The parent is shifted down $7$, so the $y$-intercept moves from $(0,0)$ to $(0,-7)$. The slope is unchanged at $1$.

Independent practice

69. a) Slope multiplied by $8$, so the line is much steeper than the parent; $y$-intercept unchanged at $(0,0)$. b) Slope unchanged at $1$; the parent is shifted up $10$, so the $y$-intercept is $(0,10)$. c) Slope $-1$: the parent is reflected across the $x$-axis with the same steepness; $y$-intercept unchanged at $(0,0)$. d) Slope halved to $\tfrac12$, so the line is less steep than the parent; shifted down $3$, so the $y$-intercept is $(0,-3)$.

70. $y = -6x$ is steeper. Steepness compares the size of the slope, ignoring its sign: $|-6| = 6$ and $|4| = 4$. The negative sign tells you that $y = -6x$ falls from left to right while $y = 4x$ rises, which is a different question from which is steeper.

71. $y = -3x - 2$

72. No. Reflecting across the $x$-axis sends each point $(x,y)$ to $(x,-y)$, and the $y$-intercept of the parent is $(0,0)$, which is sent to $(0,-0) = (0,0)$ — the origin is the one point the reflection leaves exactly where it was. Both $y = -x$ and $y = -2x$ still cross the $y$-axis at the origin.

73. Slope $1$; $y$-intercept $(0,6)$; equation $y = x + 6$; zero $-6$, from $x + 6 = 0$.

74. $x$	$-2$	$0$	$4$
    $y = x$	$-2$	$0$	$4$
    $y = x + 3$	$1$	$3$	$7$

    Every output moved up by the same $3$, which is what a vertical shift of $3$ means: the whole graph is lifted, and no output moves more than any other.

75. Changing $b$ adds the same amount to every output at once. Slope measures a difference between outputs — how much the output changes when the input increases by one unit — and adding the same number to both of two outputs leaves their difference untouched. So the graph moves up or down as a rigid piece, keeping the identical rise for every run.

76. The student read the $-4$ as a constant being added, but $-4$ is multiplying $x$, not standing alone. A shift down $4$ would be $y = x - 4$, with slope still $1$. In $y = -4x$ the slope is changed from $1$ to $-4$: the parent is reflected across the $x$-axis and made four times as steep, while the $y$-intercept stays exactly where it was, at $(0,0)$.

Exit ticket 5.5

77. Slope $7$; $y$-intercept $(0,0)$. The parent's slope is multiplied by $7$, making the line much steeper, and the $y$-intercept does not move.
78. Slope $1$; $y$-intercept $(0,-9)$. The parent is shifted down $9$; its steepness is unchanged.
79. Slope $-\tfrac12$; $y$-intercept $(0,4)$. The parent is reflected across the $x$-axis and made less steep, since $\left|-\tfrac12\right| < 1$, and then shifted up $4$.
80. A slope-changing transformation — multiplying $x$ by a number — affects $m$, and leaves $b$ alone, because multiplying the input by anything still sends $0$ to $0$. A vertical shift — adding a constant — affects $b$, and leaves $m$ alone, because it moves every output by the same amount and so changes no rise-over-run. The two dials are independent, which is why $y = mx + b$ needs exactly two numbers.

Lesson 5.6 — Three Equivalent Forms

Guided practice

81. $y = 2x - 4$ (slope-intercept), $2x - y = 4$ (standard), and $y - 2 = 2(x - 3)$ (point-slope).
82. Slope-intercept form, $y = 2x - 4$. The slope is $2$ and the $y$-intercept is $(0,-4)$, both read straight off the equation with no work.
83. Point-slope form, $y - 2 = 2(x - 3)$, names the point $(3,2)$, which is marked on the graph. The other two forms mention only the intercepts.
84. Let $y = 0$: $3x = 12$, so $x = 4$ and the $x$-intercept is $(4,0)$. Let $x = 0$: $4y = 12$, so $y = 3$ and the $y$-intercept is $(0,3)$.
85. Subtract $3x$: $4y = -3x + 12$. Divide every term by $4$: $y = -\tfrac34 x + 3$. Slope $-\tfrac34$; $y$-intercept $(0,3)$, which agrees with the substitution in item 84.
86. Slope-intercept form is $y = mx + b$, and it requires a slope $m$ to exist. The vertical line $x = 3$ has no slope, because the run between any two of its points is $0$; there is no number to put in for $m$, and no equation of the form $y = mx + b$ has $x = 3$ as its graph. In standard form it is $1x + 0y = 3$, which is legitimate because standard form allows $B = 0$.

Independent practice

87. a) $y = -2x + 7$; slope $-2$; $y$-intercept $(0,7)$ b) $x - 3y = 9$ gives $-3y = -x + 9$, then dividing by $-3$ gives $y = \tfrac13 x - 3$; slope $\tfrac13$; $y$-intercept $(0,-3)$ c) $5x - 2y = -8$ gives $-2y = -5x - 8$, then dividing by $-2$ gives $y = \tfrac52 x + 4$; slope $\tfrac52$; $y$-intercept $(0,4)$ d) $4x + 6y = 18$ gives $6y = -4x + 18$, then dividing by $6$ gives $y = -\tfrac23 x + 3$; slope $-\tfrac23$; $y$-intercept $(0,3)$
88. a) Subtract $3x$: $-3x + y = -5$; multiply by $-1$: $3x - y = 5$ b) Add $x$: $x + y = 2$ c) Multiply by $2$: $2y = x + 8$; subtract $x$: $-x + 2y = 8$; multiply by $-1$: $x - 2y = -8$ d) Multiply by $3$: $3y = -2x + 3$; add $2x$: $2x + 3y = 3$
89. a) Distribute: $y - 5 = 3x - 6$; add $5$: $y = 3x - 1$ b) Distribute: $y + 1 = -2x + 8$; subtract $1$: $y = -2x + 7$ c) Distribute: $y - 7 = \tfrac12 x + 3$; add $7$: $y = \tfrac12 x + 10$
90. Slope $-5$, the multiplier outside the parentheses. The point is $(-3, 4)$: point-slope form subtracts $x_1$, and $x + 3$ is $x - (-3)$, so $x_1 = -3$, while $y - 4$ gives $y_1 = 4$.
91. Let $y = 0$: $5x = 15$, so $x = 3$ and the $x$-intercept is $(3,0)$. Let $x = 0$: $-3y = 15$, so $y = -5$ and the $y$-intercept is $(0,-5)$.
92. Let $y = 0$: $2x = 14$, so $x = 7$ and the $x$-intercept is $(7,0)$. Let $x = 0$: $7y = 14$, so $y = 2$ and the $y$-intercept is $(0,2)$.
93. In $y = 2x - 4$: $2(5) - 4 = 10 - 4 = 6$, and $y = 6$. ✓ In $2x - y = 4$: $2(5) - 6 = 10 - 6 = 4$. ✓ In $y - 2 = 2(x-3)$: the left side is $6 - 2 = 4$ and the right side is $2(5-3) = 4$. ✓ All three are satisfied by the same point, which is what it means for them to describe the same line. (One point does not by itself prove equivalence, but combined with all three being linear equations that also share $(0,-4)$ and $(2,0)$, it is the check this chapter asks for.)
94. Let $y = 0$: $4x = 120$, so $x = 30$ and the $x$-intercept is $(30,0)$ — selling $30$ shirts and no hats raises exactly $\$120$. Let $x = 0$: $6y = 120$, so $y = 20$ and the $y$-intercept is $(0,20)$ — selling $20$ hats and no shirts also raises exactly $\$120$. The two intercepts are the two all-of-one-thing ways of hitting the goal.
95. Slope-intercept form, $y = 2.5x + 3$. The $\$2.50$ per mile is the slope, the rate at which the fare grows with distance, and the $\$3$ start charge is the $y$-intercept $(0,3)$, the fare before any distance is traveled. The two numbers in the story are exactly $m$ and $b$, so that form displays both without any conversion.
96. The student dropped the sign when moving the $y$-term. Subtracting $3x$ from both sides of $3x - y = 6$ gives $-y = -3x + 6$, and the left side is $-y$, not $y$, so both sides must still be multiplied by $-1$. That flips the sign of every term, including the constant: $y = 3x - 6$. Checking against the original with $x = 0$: the original gives $-y = 6$, so $y = -6$, which the correct form produces and the student's does not.

Exit ticket 5.6

97. $4x - y = 10$ gives $-y = -4x + 10$, then multiplying by $-1$ gives $y = 4x - 10$. Slope $4$; $y$-intercept $(0,-10)$.
98. Add $3x$ to both sides: $3x + y = 8$.
99. Distribute: $y - 6 = 4x - 4$; add $6$: $y = 4x + 2$.
100. Let $y = 0$: $6x = 30$, so $x = 5$ and the $x$-intercept is $(5,0)$. Let $x = 0$: $5y = 30$, so $y = 6$ and the $y$-intercept is $(0,6)$.

Chapter 5 Review

Part A — Domain, range, and zeros

101. Domain: all real numbers. Range: all real numbers, since the slope $-7$ is not zero. Zero: $-7x + 2 = 0$ gives $7x = 2$, so $x = \tfrac27$.
102. Domain: all real numbers. Range: the single value $9$. It has no zero — the line sits nine units above the $x$-axis at every input and never reaches height $0$.
103. No, $x = -1$ is not a function: the single input $-1$ is paired with every output at once, so it fails the vertical line test against its own graph. It has no slope, because the run between any two of its points is $0$.
104. Domain $0 \le x \le 4$; range $0 \le y \le 8$; zero $4$. In context: the pouring lasts from the start up to $4$ minutes and no longer; the amount of water in the cooler is always between $0$ and its full $8$ liters; and the zero says the cooler is empty after $4$ minutes, which is why the domain stops exactly there.
105. $45 - 3t = 0$ gives $3t = 45$, so the zero is $t = 15$: the elevator reaches the ground $15$ seconds after it begins descending. Domain $0 \le t \le 15$ seconds — the descent begins at $0$ and is over at the zero. Range $0 \le E \le 45$ meters — the elevator is somewhere between the ground and its starting height of $45$ meters.
106. A nonzero slope means the line is slanted, so it climbs (or falls) steadily and crosses the horizontal $x$-axis exactly once — it cannot cross twice, because it would have to turn around to come back, and a line does not turn. Algebraically, $mx + b = 0$ has exactly one solution, $x = -\tfrac{b}{m}$, precisely because $m \ne 0$ makes the division possible. A horizontal line other than $y = 0$ has no zero: its slope is $0$, it never meets the $x$-axis, and $0 \cdot x + b = 0$ has no solution when $b \ne 0$.

Part B — Slope and intercepts

107. Slope $2$; $y$-intercept $(0,-6)$; $x$-intercept from $2x - 6 = 0$, so $x = 3$, giving $(3,0)$.
108. $m = \dfrac{-3 - 9}{2 - (-4)} = \dfrac{-12}{6} = -2$
109. $y$-intercept $(0,15)$; $x$-intercept from $-3x + 15 = 0$, so $3x = 15$ and $x = 5$, giving $(5,0)$.
110. The slope $30$ is a rate of $30$ gallons per minute: the pool gains $30$ gallons of water each minute the hose runs. The $y$-intercept $(0,150)$ is the starting amount: the pool already held $150$ gallons when the filling began, so it was not empty at the start.
111. Slope $\tfrac34$, from the dashed triangle labeled rise $3$ and run $4$. $y$-intercept $(0,-3)$, from the marked point where the line crosses the $y$-axis. $x$-intercept $(4,0)$, from the marked point where the line crosses the $x$-axis. Zero $4$, which is the $x$-coordinate of that same $x$-intercept — the figure's caption names it, and no new part of the picture is needed for it.
112. The student gave the $y$-intercept instead. An $x$-intercept has $y$-coordinate $0$, not $x$-coordinate $0$. Correctly: the $y$-intercept is $(0,8)$, and the $x$-intercept comes from $4x + 8 = 0$, so $4x = -8$ and $x = -2$, giving $(-2, 0)$.

Part C — The parent function and transformations

113. The slope is multiplied by $6$, from $1$ to $6$, so the line is much steeper than the parent. The graph is shifted down $1$, so the $y$-intercept moves from $(0,0)$ to $(0,-1)$.
114. The slope becomes $-1$: the parent is reflected across the $x$-axis, with the same steepness, so the line falls where the parent rises. The graph is shifted up $5$, so the $y$-intercept moves to $(0,5)$.
115. Left panel: the slope changed — $y = 3x$ is steeper and $y = \tfrac13 x$ is less steep than the parent — and the $y$-intercept stayed at $(0,0)$. Middle panel: the direction changed — a negative slope reflects the parent across the $x$-axis, so $y = -x$ and $y = -2x$ fall instead of rise — and the $y$-intercept stayed at $(0,0)$. Right panel: the $y$-intercept changed, to $(0,3)$ and $(0,-4)$, and the slope stayed at $1$.
116. Any two functions of the form $y = mx$ with different values of $|m|$. For example $y = 2x$ and $y = \tfrac14 x$: both pass through $(0,0)$, because multiplying the input never moves the origin, and $y = 2x$ is the steeper of the two, since $|2| > \left|\tfrac14\right|$.
117. Every linear function can be written $y = mx + b$, and that expression is built out of the parent $y = x$ by exactly two moves: multiply the input by $m$, which sets the slope, and add $b$, which sets the $y$-intercept. Since $m$ and $b$ can be any numbers, every linear function is reachable, so every one of them is a transformation of $y = x$. The lines this does not cover are the vertical lines $x = a$: they have no slope, so there is no $m$ to choose, and they are not functions at all, since one input carries every output.

Part D — Three equivalent forms

118. Subtract $x$: $4y = -x + 8$; divide every term by $4$: $y = -\tfrac14 x + 2$. Slope $-\tfrac14$; $y$-intercept $(0,2)$.
119. Subtract $5x$: $-5x + y = -3$; multiply by $-1$: $5x - y = 3$.
120. Distribute: $y + 2 = -3x + 3$; subtract $2$: $y = -3x + 1$.
121. Let $y = 0$: $4x = 20$, so $x = 5$ and the $x$-intercept is $(5,0)$. Let $x = 0$: $-5y = 20$, so $y = -4$ and the $y$-intercept is $(0,-4)$.
122. Standard form alone can write $x = 3$, as $1x + 0y = 3$; slope-intercept form cannot, because $x = 3$ has no slope and there is no $m$ to supply, and point-slope form cannot either, for the same reason. All three can write $y = -2$: it is $0x + 1y = -2$ in standard form, $y = 0x - 2$ in slope-intercept form, and $y - (-2) = 0(x - x_1)$ in point-slope form. The difference is that a horizontal line has a slope — it is $0$ — while a vertical line has none, and the two forms built around $m$ can only describe lines that have one. Standard form does not mention slope at all, which is why it covers every line in the plane.

Part E — Mixed application

123. Slope $1.75$; $y$-intercept $(0,4)$; zero from $1.75m + 4 = 0$, so $1.75m = -4$ and $m = -\tfrac{16}{7}$, about $-2.29$. The slope means the fare grows by $\$1.75$ for each additional mile driven. The $y$-intercept means a ride of zero miles still costs $\$4$ — the base charge for getting in the car. The zero has no meaning in this situation because it is a negative number of miles, and the realistic domain of the model is $m \ge 0$. The graph would have to cross the $x$-axis to the left of the vertical axis, in a region the story never enters: with a positive base charge and a positive rate, the fare is never $\$0$ for any real ride.
124. Let $y = 0$: $5x = 200$, so $x = 40$ and the $x$-intercept is $(40,0)$ — selling $40$ tickets and no programs raises exactly $\$200$. Let $x = 0$: $8y = 200$, so $y = 25$ and the $y$-intercept is $(0,25)$ — selling $25$ programs and no tickets also raises $\$200$. Converting: $8y = -5x + 200$, so $y = -\tfrac58 x + 25$. That form reveals the rate of trade between the two items, which standard form kept hidden: the slope $-\tfrac58$ says that every extra ticket sold reduces the number of programs still needed by $\tfrac58$ of a program — equivalently, every $8$ extra tickets let the club sell $5$ fewer programs and still reach $\$200$.

Workbook-only items

Page 2, fill in the blanks. For a linear function with a nonzero slope, the domain is all real numbers and the range is all real numbers. Arrowheads say the graph continues past the edge of the grid.

Page 3, the two special lines. A horizontal line has slope $m = \mathbf{0}$; its range is the single value $b$ (here, $3$), and it has no zero. A vertical line $x = a$ is not a function, and its slope does not exist.

Page 4, three questions. Can the input be negative? Can the input be a fraction? Where does the story end?

Page 7, complete the formulas. $m = \dfrac{\textbf{rise}}{\textbf{run}} = \dfrac{y_2 - y_1}{\mathbf{x_2 - x_1}}$. In $y = mx + b$ the slope is $\mathbf{m}$ and the $y$-intercept is the point $\mathbf{(0,b)}$. A positive slope rises from left to right; a negative slope falls.

Page 9, intercept frame. To find the $x$-intercept, set $y$ (the output) equal to $0$ and solve for $x$.

Page 12, three names frame. The zero of $f$ is the input with output $\mathbf{0}$; the $x$-intercept is the point $\mathbf{(x, 0)}$; the solution of the equation $\mathbf{mx + b = 0}$. A zero is a number; an $x$-intercept is a point.

Page 15, parent frame. In $y = x$, each output equals its own input. Written out, the parent is $y = \mathbf{1}x + \mathbf{0}$.

Page 17, two dials. Multiplying $x$ by a number changes the slope and leaves the $y$-intercept at $(0,0)$. Adding a constant changes the $y$-intercept and leaves the slope at $\mathbf{1}$. If $|m| > 1$ the line is steeper than the parent; if $0 < |m| < 1$ it is less steep. A negative slope reflects the parent across the $x$-axis.

Page 21, forms table.

Page 22, standard-form frame. In $Ax + By = C$, let $y = 0$ to find the $x$-intercept and let $x = 0$ to find the $y$-intercept.

Page 32, blank grids. Any sketch a teacher assigns. The expected conventions are the two printed on the page: a linear function with no restricted domain is drawn all the way across the grid with an arrowhead on each end, because its domain is all real numbers; a linear function modeling a situation with a beginning and an end is drawn as a segment with a filled-in point at each endpoint, because both endpoints are real moments in the story and are included in the domain. A student checking work on these grids should be able to find the two intercepts as grid corners and to count a slope triangle between them.