Chapter 7 — Graphing and Evaluating Linear Functions

Standard: A.F.1 (f, g, h)

A.F.1 — verbatim. The student will investigate, analyze, and compare linear functions algebraically and graphically, and model linear relationships. Students will demonstrate the following Knowledge and Skills: f) Graph a linear function in two variables, with and without the use of technology, including those that can represent contextual situations. g) For any value, $x$, in the domain of $f$, determine $f(x)$, and determine $x$ given any value $f(x)$ in the range of $f$, given an algebraic or graphical representation of a linear function. h) Compare and contrast the characteristics of linear functions represented algebraically, graphically, in tables, and in contextual situations.

By the end of this chapter you will be able to:

• Graph a linear function by hand from slope-intercept form, by plotting the $y$-intercept and stepping off the slope in either direction (A.F.1f)
• Graph a linear function by hand from its two intercepts, and from a table of inputs you choose on purpose (A.F.1f)
• Graph a horizontal line $y = c$ and a vertical line $x = a$ without stepping off any slope (A.F.1f)
• Graph a contextual linear function on axes whose two scales differ, and say what one cell on each axis is worth (A.F.1f)
• Graph a linear function with technology, choose a viewing window from what you already know about the function, and use the table feature (A.F.1f)
• Given $x$, determine $f(x)$ — from the equation and from the graph (A.F.1g)
• Given $f(x)$, recover the $x$ that produced it — from the equation and from the graph (A.F.1g)
• Compare and contrast two linear functions presented in any of four ways — algebraically, graphically, in a table, or in a context — and say which representation answers which question fastest (A.F.1h)

Lessons: 7.1 Graphing by Hand from Slope-Intercept Form · 7.2 Intercepts, Tables, and the Two Special Lines · 7.3 Graphing with Technology · 7.4 Determining $f(x)$, and Recovering $x$ · 7.5 Comparing the Four Representations

Why this chapter matters. Chapter 5 read a linear function, and Chapter 6 wrote one. This chapter makes one usable. A model nobody can draw and nobody can query is just an ornament: the two questions people actually bring to a linear model are "what will it be at this input?" and "what input gets me to this output?", and both of them are in bullet g. Bullet f gives you the picture that answers them at a glance, by hand when you have paper and with technology when the numbers are ugly. Bullet h then steps back and asks the question that closes out the whole linear sequence: given the same relationship written four different ways, what does each way show you that the others hide?

Scope note. This chapter draws and queries linear functions. Identifying and interpreting domain, range, zeros, slope, and intercepts, and converting among slope-intercept, standard, and point-slope form, are A.F.1 a, b, and c, in Chapter 5; this chapter uses all of them constantly and says so when it does. Writing the equation of a line — from a graph, from two points, from a slope and a point, and parallel or perpendicular to a given line — is A.F.1 d and e, in Chapter 6; when this chapter hands you an equation, that equation is a given, not something to be derived. Bullet h is taught here rather than earlier because it is the capstone of the linear sequence: comparing across representations is only possible once you can produce all four of them, and Chapter 5, Chapter 6, and Lessons 7.1 through 7.4 are what make that true. Systems of two linear equations are Chapter 8, so when Lesson 7.5 compares two models on one grid it reads the crossing point off the picture and confirms it by solving a one-variable equation, the way Chapter 2 taught.

Conventions this chapter fixes.

• A graph drawn on a bare grid gets arrowheads on both ends unless a context restricts the domain, in which case it is a segment with its endpoints marked, exactly as in Chapter 5.
• Slope is a fraction, always. An integer slope $m$ is read as $\tfrac{m}{1}$ before stepping, because the denominator is the run and there has to be one.
• Notation for the two directions of bullet g. "Find $f(3)$" means substitute and simplify. "Find $x$ when $f(x) = 3$" means set the rule equal to $3$ and solve. Both name the same point on the graph from opposite ends.
• A viewing window is reported as the four numbers a calculator asks for: the smallest and largest $x$, and the smallest and largest $y$, together with the scale — what one tick or one cell is worth on each axis.
• Technology is an instrument of verification, not a shortcut. Algebra 1 has no no-calculator standards, and the Desmos Virginia Graphing Calculator is available for the entire End-of-Course test. This volume therefore asks for the algebraic result and its graphical confirmation. When the two disagree, that is information: one of them is wrong, and you are not finished until you know which.
• A calculator reports decimals; a zero may be exact. If the screen says $3.3333333$ and the algebra says $\tfrac{10}{3}$, the exact answer is $\tfrac{10}{3}$ and the screen is rounding.
• Item numbering runs straight through the chapter, from 1 in Lesson 7.1 to 120 at the end of the review. It does not restart at each lesson.

Lesson 7.1 — Graphing by Hand from Slope-Intercept Form

Why "without technology" is a real skill

A.F.1f asks for graphing with and without the use of technology, and it names them in that order for a reason: the by-hand method is not a fallback for when the batteries die. It is how you know what the screen should look like before you look at it. A student who can step off a slope knows, without pressing anything, that $y = 25x - 140$ is steep and crosses the $y$-axis far below the origin — and so is not surprised when the standard window shows almost nothing.

The by-hand method takes a function in slope-intercept form $y = mx + b$, the form Chapter 5 named and Chapter 6 reported every answer in.

The procedure

1. Plot the $y$-intercept. It is $(0, b)$, and it is free — no arithmetic at all.
2. Write the slope as a fraction. The denominator is the run, the numerator is the rise.
3. Step off the slope from the point you just plotted, and put a dot where you land.
4. Step again to get a third point. Two points determine a line; the third one catches an error before you draw.
5. Draw the line through your points, past all of them, with an arrowhead on each end.

The figure works $f(x) = \tfrac23 x - 4$ all the way through. The $y$-intercept is $(0, -4)$. The slope $\tfrac23$ means right 3, up 2, which lands on $(3, -2)$; stepping again lands on $(6, 0)$.

Check the third point against the rule, which is the whole point of plotting it:

$$f(6) = \tfrac23(6) - 4 = 4 - 4 = 0 \quad \checkmark$$

Notice that $(6, 0)$ is the $x$-intercept, so a graph built this way often hands you the zero for free.

Stepping in either direction

A slope can be walked backwards. If $\tfrac{\text{rise}}{\text{run}}$ takes you right and up, then reversing both signs takes you left and down, and you stay on the same line.

That matters most for a negative slope, where the two directions look completely different on the page.

The figure shows $h(x) = -2x + 3$. Its slope is the integer $-2$, so write it as a fraction first:

$$m = -2 = \frac{-2}{1}$$

The run is $1$ and the rise is $-2$: right 1, down 2, from $(0,3)$ to $(1,1)$ to $(2,-1)$.

Reversing both gives a run of $-1$ and a rise of $+2$: left 1, up 2, from $(0,3)$ to $(-1,5)$. Substituting confirms it: $h(-1) = -2(-1) + 3 = 5$.

Two habits fall out of this.

• Put the negative sign on the rise, not the run. Writing $-2$ as $\tfrac{2}{-1}$ is legal but invites a sign error; $\tfrac{-2}{1}$ says "down 2" out loud.
• Step both ways when the intercept is near an edge. If $b = -8$ on a grid that stops at $-8$, you cannot step down. Step up and left instead.

Where students lose points

• Stepping the fraction upside down. For $m = \tfrac25$, the run is $5$, not $2$. The denominator is always the horizontal move. If you find yourself going up $5$ and right $2$, you have graphed $y = \tfrac52 x + b$ instead.
• Forgetting that $b$ is a $y$-coordinate. The first point of $y = 3x - 4$ is $(0,-4)$, on the $y$-axis. It is not $(-4, 0)$, and it is certainly not $(0,3)$.
• Drawing a segment when the function has no restriction. With no context, the domain is all real numbers, so the ink gets arrowheads.

Worked examples

Example 1 — An integer slope

Graph $y = 3x - 2$ by hand. Name the first point and two more.

The $y$-intercept is $(0,-2)$. The slope $3 = \tfrac31$ means right 1, up 3.

Answer: Plot $(0,-2)$; step to $(1,1)$ and $(2,4)$; draw with arrowheads.

Example 2 — A fractional slope

Graph $y = \tfrac{3}{5}x + 1$. Name the first point and two more.

The $y$-intercept is $(0,1)$, and $\tfrac35$ means right 5, up 3.

Answer: $(0,1)$, then $(5,4)$, then $(10,7)$. Going the other way, $(-5,-2)$ is also on the line and fits on a smaller grid.

Example 3 — A negative slope, stepped both ways

Graph $y = -x + 5$. Give one point on each side of the $y$-intercept.

The $y$-intercept is $(0,5)$ and $m = \tfrac{-1}{1}$.

Answer: Right 1, down 1 gives $(1,4)$. Left 1, up 1 gives $(-1,6)$.

Example 4 — Choosing the direction to fit the grid

The grid runs from $-6$ to $6$ on both axes. Graph $y = \tfrac14 x + 5$.

Stepping right 4 and up 1 from $(0,5)$ lands on $(4,6)$, still on the grid, but a second step leaves the top.

Answer: Plot $(0,5)$, step right to $(4,6)$, and step left to $(-4,4)$ and $(-8,3)$ — the left-hand steps stay inside the window far longer.

Example 5 — Catching an error with the third point

A student graphs $y = \tfrac23 x - 4$ and plots $(0,-4)$, $(3,-2)$, and $(6,1)$. Is anything wrong?

Check the third point in the rule: $\tfrac23(6) - 4 = 0$, not $1$.

Answer: The third point should be $(6,0)$. The student stepped up $3$ instead of $2$ on the second step — exactly the mistake a third point exists to catch.

Guided practice

1. Use the figure of $f(x) = \tfrac23 x - 4$. Name the point plotted first, the slope, and the two other points the steps produce.
2. In that same figure, what do "up 2" and "right 3" refer to, and which one is the denominator of the slope?
3. In that same figure, verify $f(6) = 0$ algebraically, and say which intercept that point is.
4. Use the figure of $h(x) = -2x + 3$. Write the slope as a fraction and state the step it names.
5. In that same figure, what point does "left 1, up 2" produce? Verify it by substitution.
6. Graph $y = 3x - 2$ by hand. Give the first point plotted and two more points.

Independent practice

7. For each function, give the point you plot first and the step you take. a) $y = 4x + 1$ b) $y = -x + 5$ c) $y = \tfrac35 x - 2$ d) $y = -\tfrac23 x + 4$
8. Graph $y = \tfrac12 x - 3$ on Grid A. List three points with integer coordinates.
9. Graph $y = -3x + 6$ on Grid B. List three points with integer coordinates.
10. Error analysis. A student graphs $y = \tfrac25 x + 1$ by plotting $(0,1)$ and then going up $5$ and right $2$. Identify the error, name the function that student actually graphed, and give the correct step.
11. List four points with integer coordinates on $y = -\tfrac34 x + 2$, using inputs on both sides of the $y$-axis.
12. Reasoning. Explain why stepping left and up by the reversed slope lands on the same line as stepping right and down. Use $h(x) = -2x + 3$ in your explanation.
13. Application. An office printer starts a shift with $500$ sheets of paper and uses $25$ sheets per job, so $P(j) = 500 - 25j$. Name the point you would plot first and the step you would take, then explain why a grid of one sheet per cell is a bad choice.
14. Graph $y = 2x$ and $y = 2x + 3$ by hand on the same grid. Name the first point of each, and say what is the same about the two graphs and what is different.
15. Application. A phone battery falls from full according to $B(t) = 100 - 8t$ percent after $t$ hours. Give the first point and the step, and find how far right you must travel before the graph reaches $0$.
16. Error analysis. A student says the first point of $y = 3x - 4$ is $(0,3)$, "because $3$ comes first." Identify the error and give the correct first point and step.

Exit ticket 7.1

17. Give the first point plotted and the step for $y = -\tfrac43 x + 5$.
18. List three points with integer coordinates on $y = \tfrac13 x - 1$.
19. Explain why two points are enough to draw a line, and why plotting a third is still worth the time.
20. Graph $y = -2x + 1$ by hand. Give the $y$-intercept and two more points.

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

Three by-hand methods, not one

Slope stepping is the default, but it is not always the cheapest route. Two other by-hand methods are faster in the situations they suit, and two lines need no method at all.

Graphing from the two intercepts

Chapter 5 showed that standard form makes both intercepts nearly free, because setting one variable to zero deletes a whole term. Two points are all a line needs, so that is a complete graphing method.

For $3x - 4y = 12$:

1. Let $x = 0$. Then $-4y = 12$, so $y = -3$. Plot $(0,-3)$.
2. Let $y = 0$. Then $3x = 12$, so $x = 4$. Plot $(4,0)$.
3. Draw the line through the two points, with arrowheads.

No solving for $y$, no slope, no fractions along the way. (If you want the slope anyway, converting gives $y = \tfrac34 x - 3$, and the graph confirms a rise of $3$ over a run of $4$ between the two plotted points.)

The method has one blind spot worth naming: a line through the origin has only one intercept, because both are the same point $(0,0)$. For $y = \tfrac52 x$ you must find a second point some other way — substitute any convenient input.

Graphing from a table you choose

A table is the most general method, and its whole art is in choosing the inputs.

The figure graphs $f(x) = \tfrac12 x + 1$ from a table:

Every output is a whole number, so every point lands exactly on a grid corner. That did not happen by luck. Choosing $x = 1, 3, 5$ would have produced $1.5$, $2.5$, $3.5$ — points you have to eyeball between gridlines, which is how a hand-drawn line ends up crooked.

The rule for choosing inputs. Pick inputs that are multiples of the denominator of the slope. For a slope of $\tfrac12$, use even numbers; for $\tfrac23$, use multiples of $3$; for $-\tfrac25$, use multiples of $5$. And always include $0$ if you can, because it hands you the $y$-intercept.

The two lines that need no stepping

• $y = -3$ is every point whose height is $-3$. Find $-3$ on the $y$-axis and rule a horizontal line. Its slope is $0$, it is a function, and it has no zero.
• $x = 4$ is every point whose first coordinate is $4$. Find $4$ on the $x$-axis and rule a vertical line. It has no slope and is not a function, so there is no $f(x)$ to evaluate on it — which is why Lesson 7.4 never asks you to.

Both were written in Chapter 6 as $y = c$ and $x = a$. Here they are drawn.

Graphing a context: choosing a scale

A bare-grid graph uses one unit per cell on both axes. A context almost never can, because the two axes count different things.

A bike shop charges a $\$20$ flat fee plus $\$5$ per hour, so the cost of an $h$-hour rental is

$$C(h) = 5h + 20$$

Rentals run up to $8$ hours. Drawing that with one dollar per cell would need a grid $60$ cells tall.

Here is the routine.

1. Find the range of inputs the story allows. Here $0 \le h \le 8$, so the horizontal axis needs $8$ cells at $1$ hour each.
2. Find the largest and smallest outputs those inputs produce. $C(0) = 20$ and $C(8) = 60$, so the vertical axis must reach at least $60$.
3. Pick a scale that makes the picture fit and keeps the plotted points on corners. At $5$ dollars per cell, $60$ dollars is $12$ cells and every hour's cost — $20$, $25$, $30$, and so on — lands exactly on a gridline.
4. Label both axes with the quantity and its unit, and say what one cell is worth. An unlabeled context graph is unreadable.
5. Draw a segment, not a line. The domain is $0 \le h \le 8$, so the graph stops at both ends, with the endpoints marked, exactly as Chapter 5 taught.

A scale is not a matter of taste. Choosing $7$ dollars per cell would put $C(1) = 25$ somewhere between two gridlines, and every point after it too.

Worked examples

Example 1 — From intercepts

Graph $2x + 5y = 10$ by hand.

Let $x = 0$: $5y = 10$, so $y = 2$. Let $y = 0$: $2x = 10$, so $x = 5$.

Answer: Plot $(0,2)$ and $(5,0)$ and draw the line.

Example 2 — Choosing inputs for a table

Build a four-row table for $f(x) = \tfrac23 x + 2$ with whole-number outputs.

The denominator of the slope is $3$, so use multiples of $3$.

Answer: $x = -3, 0, 3, 6$ gives $f(x) = 0, 2, 4, 6$.

Example 3 — A line through the origin

Why does the intercept method fail for $y = \tfrac52 x$, and what do you do instead?

Setting $x = 0$ gives $y = 0$, and setting $y = 0$ gives $x = 0$ — the same point twice.

Answer: Both intercepts are $(0,0)$, so you have one point, not two. Substitute a convenient input instead: $x = 2$ gives $y = 5$, so plot $(2,5)$.

Example 4 — A horizontal and a vertical line

Graph $y = 6$ and $x = -5$, and say which is a function.

Answer: $y = 6$ is horizontal through $6$ on the $y$-axis; slope $0$; it is a function. $x = -5$ is vertical through $-5$ on the $x$-axis; no slope; it is not a function.

Example 5 — Choosing a scale

A pool fills according to $G(t) = 12t + 30$ gallons after $t$ minutes, for $0 \le t \le 20$. Choose scales and name the endpoints.

$G(0) = 30$ and $G(20) = 12(20) + 30 = 270$.

Answer: Horizontal axis $0$ to $20$ minutes at $2$ minutes per cell; vertical axis $0$ to $280$ gallons at $20$ gallons per cell. The segment runs from $(0,30)$ to $(20,270)$.

Guided practice

21. Use the figure of $3x - 4y = 12$. Name the two substitutions shown and the two points they produce.
22. Convert $3x - 4y = 12$ to slope-intercept form, and confirm that the slope agrees with the two points plotted in that figure.
23. Use the figure of $f(x) = \tfrac12 x + 1$. Which five inputs were chosen, and why those?
24. In that same figure, what would have gone wrong with the inputs $1$, $3$, and $5$?
25. Use the figure of the two special lines. Describe how you graph $y = -3$ and how you graph $x = 4$ without stepping off any slope, and say which of the two is a function.
26. Use the bike-rental figure. State what one cell across and one cell up is worth, and give the cost of a $4$-hour rental.

Independent practice

27. Graph each by finding both intercepts. Give the two points. a) $2x + 5y = 10$ b) $3x - y = 6$ c) $x - 4y = 8$ d) $6x + 3y = -12$
28. Build a four-row table for $f(x) = \tfrac23 x + 2$ whose outputs are all whole numbers, and say how you chose the inputs.
29. Build a four-row table for $f(x) = -\tfrac14 x + 5$ whose outputs are all whole numbers.
30. Graph $y = -1$ and $x = 3$ on one grid. For each, give the slope if it has one and say whether it is a function.
31. Application. Use the bike-rental function $C(h) = 5h + 20$ from the figure. Find $C(3)$ and $C(6)$, and explain why $5$ dollars per cell is a better vertical scale than $4$ dollars per cell.
32. Application. A car rental charges a $\$45$ fee plus $\$0.30$ per mile, so $D(m) = 0.30m + 45$ for $0 \le m \le 200$. Choose a scale for each axis, say what one cell is worth, and give the two endpoints of the segment.
33. Reasoning. For $4x - 5y = 20$, is it faster to graph from the intercepts or to convert to slope-intercept form first? Give both intercepts, and justify your choice.
34. Error analysis. A student graphs $2x + 3y = 12$ by plotting $(0,3)$ and stepping right $1$ and up $2$, "because the slope is $2$ and the intercept is $3$." Identify the error and give a correct graph as two points.
35. Application. A booster club sells wristbands at $\$8$ and shirts at $\$12$ toward a $\$240$ goal, so $8x + 12y = 240$. Graph it from its intercepts, and say what each intercept means about the fundraiser.
36. Application. A pool holds $400$ gallons and drains at $25$ gallons per minute, so $W(t) = 400 - 25t$ for $0 \le t \le 16$. Choose a scale for each axis and say what one cell is worth on each.

Exit ticket 7.2

37. Graph $5x + 2y = 20$ from its intercepts. Name both points.
38. Give four inputs for $f(x) = \tfrac34 x - 1$ that keep every output a whole number, and give the outputs.
39. Graph $y = 6$ and $x = -5$. Say which is a function and give the slope of each, if it has one.
40. Application. A pool fills according to $G(t) = 12t + 30$ gallons after $t$ minutes, for $0 \le t \le 20$. Choose scales for both axes and give the two endpoints of the segment.

Lesson 7.3 — Graphing with Technology

What technology is for

A.F.1f names technology explicitly, and this volume takes that seriously in one specific way: the calculator confirms, it does not decide. You produce a result algebraically, you produce it again on the screen, and you compare. The habit is the same one Chapter 6 built when it checked every equation it wrote.

Algebra 1 has no no-calculator standards, and the Desmos Virginia Graphing Calculator is available for the entire End-of-Course test — so the question is never whether to use it, only how.

The one thing the calculator gets wrong for you: the window

Enter a function and press graph and you get something. Whether that something is informative is entirely up to the window you chose.

The standard window runs from $-10$ to $10$ on both axes. In it, $f(x) = 25x - 140$ is a near-vertical stroke through one corner. Neither intercept is on the screen, and a student who trusts the picture might conclude the function has no $y$-intercept at all.

The fix is not to hunt for a better window by trial and error. It is to work out, on paper, what the window has to contain.

1. Find the $y$-intercept. Here $b = -140$, so the window must reach at least that far down.
2. Find the zero. $25x - 140 = 0$ gives $x = \tfrac{140}{25} = 5.6$, so the window must reach at least $x = 5.6$.
3. Set the window generously around both, and choose a scale for each axis the way Lesson 7.2 did.

For this function, $x$ from $0$ to $10$ with a scale of $1$, and $y$ from $-150$ to $100$ with a scale of $25$, shows the whole story. Both intercepts are on screen, and the tick marks are worth reading.

You will notice that step 1 and step 2 are Chapter 5's work. You need the algebra before the technology can help you, which is exactly why this chapter teaches the by-hand methods first.

The table feature

Every graphing calculator will also print a table of values, and it is the fastest way to bracket a zero when the zero is not a whole number.

For $f(x) = 25x - 140$, a table stepping by $1$ from $x = 0$ reaches

The output changes sign between $x = 5$ and $x = 6$, so the zero is between them — and $5.6$ is. A table also lets you check a single evaluation instantly, which is the verification half of Lesson 7.4.

Reading a value off the screen

Calculators offer a "value" or "trace" command that reports $f(x)$ at an input you type, and a "zero" or "root" command that reports where the graph crosses the $x$-axis. Both are verification tools, and both come with a caution.

A calculator reports decimals. Some answers are not decimals. Ask a calculator for the zero of $f(x) = 3x - 10$ and it will say something like $3.3333333$. The exact zero is $\tfrac{10}{3}$. The screen is rounding; report the exact value, and use the decimal only to confirm it. Likewise a zero displayed as $2.9999999$ for $f(x) = 5x - 15$ is the number $3$, and $3$ is what you write down.

When the two disagree

This is the rule that makes verification worth doing.

If your algebra and your graph disagree, one of them is wrong, and you are not finished. Do not average them, and do not pick the one you like. Find the error. The usual suspects, in order:

1. The function was typed wrong. A missing parenthesis turns $\tfrac{x+1}{2}$ into $x + \tfrac12$.
2. The window is hiding something. The feature you are looking for is off screen.
3. A sign was dropped in the algebra. Re-solve, slowly.

If you graph $y = 2x + 5$ and the screen shows a line falling from left to right, you have not discovered a new fact about slopes — you typed $-2x + 5$.

Worked examples

Example 1 — Setting a window from the algebra

Choose a viewing window for $f(x) = 12x + 96$.

The $y$-intercept is $(0,96)$. The zero: $12x + 96 = 0$ gives $x = -8$.

Answer: $x$ from $-12$ to $4$ with a scale of $2$; $y$ from $-20$ to $120$ with a scale of $20$. Both intercepts are then on screen.

Example 2 — Bracketing a zero with a table

Use a table for $f(x) = -6x + 15$ at $x = 0, 1, 2, 3, 4$ to locate the zero.

The outputs are $15$, $9$, $3$, $-3$, $-9$.

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

Example 3 — Algebra first, graph second

Solve $4x - 7 = 13$, then describe the graphical confirmation.

$4x = 20$, so $x = 5$.

Answer: $x = 5$. To confirm, graph $f(x) = 4x - 7$ and check that the point $(5, 13)$ is on the line — for instance with the table feature, which should print $13$ beside the input $5$.

Example 4 — An invisible intercept

A student graphs $f(x) = 0.1x - 4$ in the standard window and concludes it has no $x$-intercept. What went wrong?

$0.1x - 4 = 0$ gives $x = 40$, far outside a window that stops at $10$.

Answer: The zero is $40$; the window was too narrow. A window with $x$ from $-10$ to $50$ shows it.

Example 5 — An exact answer behind a decimal

A calculator reports the zero of $f(x) = 3x - 10$ as $3.3333333$. What is the zero?

$3x - 10 = 0$ gives $x = \tfrac{10}{3}$.

Answer: The exact zero is $\tfrac{10}{3}$, which is about $3.33$. The screen is showing a rounded decimal, not a different number.

Guided practice

41. Use the left panel of the window figure. Explain why the standard window is useless for $f(x) = 25x - 140$.
42. Find the $y$-intercept and the zero of $f(x) = 25x - 140$ algebraically, and say how each one determined a number in the window on the right.
43. Use the right panel of that figure. What is one cell worth on each axis?
44. Choose a viewing window that shows both intercepts of $f(x) = 12x + 96$, and give the two intercepts.
45. A calculator table for $f(x) = 25x - 140$ steps by $1$. What does it print at $x = 5$ and at $x = 6$, and what does the change tell you?
46. Explain in one sentence what a graph can verify about an algebraic result, and one thing it cannot establish exactly.

Independent practice

47. Choose a viewing window for each function, giving the smallest and largest $x$, the smallest and largest $y$, and a scale for each axis. State the intercepts you used to decide. a) $f(x) = 3x - 45$ b) $f(x) = -\tfrac12 x + 8$ c) $f(x) = 200x + 100$ d) $f(x) = x - 2$
48. Application. A student saves for tuition according to $T(w) = 35w + 400$ dollars after $w$ weeks, for $0 \le w \le 52$. Choose a window, and find $T(52)$.
49. You graph $y = 2x + 5$ and the screen shows a line through $(0,5)$ falling from left to right. Name the most likely cause and say how to check it.
50. Solve $4x - 7 = 13$ algebraically, then describe exactly what you would look for on a graph of $f(x) = 4x - 7$ to confirm your solution.
51. Error analysis. A student graphs $f(x) = 0.1x - 4$ in the standard window and writes "no $x$-intercept." Identify the error and give the zero.
52. Build the calculator table for $f(x) = -6x + 15$ at $x = 0, 1, 2, 3, 4$. Say between which two inputs the zero lies, then find the zero exactly.
53. Reasoning. Two students graph the same function on the same model of calculator and get pictures that look nothing alike. Give the most likely explanation, and say what the two students should compare first.
54. Application. A moving service charges $C(m) = 1.2m + 30$ dollars for $m$ miles. Find the mileage that costs $\$90$ algebraically, then describe the graphical check.
55. Reasoning. Explain why finding a good viewing window requires knowing something about the function first, and name the two numbers you should compute before touching the calculator.
56. A calculator reports the zero of $f(x) = 3x - 10$ as $3.3333333$. Give the exact zero, and explain the difference between what the screen shows and what you should write.

Exit ticket 7.3

57. Choose a viewing window for $f(x) = -15x + 120$, and give the intercepts that justified it.
58. Find $f(7)$ for $f(x) = -15x + 120$ algebraically, and describe the graphical check.
59. State the rule this chapter uses when an algebraic result and a graph disagree, and list two things to check first.
60. Error analysis. A calculator reports the zero of $f(x) = 5x - 15$ as $2.9999999$, and a student writes that as the answer. Identify the problem and give the exact zero.

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

Two questions, one point

A.F.1g asks two things, and they are opposite ends of the same fact.

• Given $x$, determine $f(x)$. You know the input; you want the output.
• Given $f(x)$, determine $x$. You know the output; you want the input that produced it.

Both name a single point $(x, f(x))$ on the graph. The difference is which coordinate you were handed.

The standard requires both directions from both representations, so all four cells of that table get worked below.

Given $x$, determine $f(x)$

Algebraically, substitute the input everywhere $x$ appears and simplify. For $f(x) = 2x - 3$:

$$f(4) = 2(4) - 3 = 8 - 3 = 5$$

Two cautions. Put the input in parentheses, especially when it is negative: $f(-4) = 2(-4) - 3 = -11$, and a missing pair of parentheses is where the sign gets lost. And an input need not be an integer: $f\!\left(\tfrac13\right)$ is a perfectly legal question, and Chapter 1 taught you to evaluate it.

Graphically, walk the path.

1. Find the input on the $x$-axis. Here, $4$.
2. Travel vertically to the line. Up, if the line is above the axis there; down, if it is below.
3. Travel horizontally to the $y$-axis and read the height. Here, $5$.

So $f(4) = 5$, which is what the algebra said. Two methods, one answer — that is the verification habit, done without a calculator.

Given $f(x)$, determine $x$

Algebraically, this is an equation to solve — the same multi-step linear equation Chapter 2 taught. Set the rule equal to the given output.

For $f(x) = 2x - 3$, find $x$ when $f(x) = -7$:

$$2x - 3 = -7 \quad\Rightarrow\quad 2x = -4 \quad\Rightarrow\quad x = -2$$

Graphically, walk the same path backwards.

1. Find the output on the $y$-axis. Here, $-7$.
2. Travel horizontally to the line.
3. Travel vertically to the $x$-axis and read the input. Here, $-2$.

The two figures show the same function in the same window with the same kind of dashed path. The only difference is which axis you start from.

The mistake this lesson exists to prevent

Asked to find $x$ when $f(x) = 10$ for $f(x) = 2x + 4$, a great many students compute

$$f(10) = 2(10) + 4 = 24$$

and report $24$. That answers the other question. The given number $10$ is an output, so it belongs on the right-hand side of an equation, not inside the parentheses:

$$2x + 4 = 10 \quad\Rightarrow\quad 2x = 6 \quad\Rightarrow\quad x = 3$$

A quick reading test: the number inside $f(\ \ )$ is always an input. If the problem writes $f(x) = 10$, the $10$ is outside the parentheses and is therefore an output.

The zero is one instance of this

Setting $f(x) = 0$ and solving is the second direction with the specific output $0$. So Chapter 5's zero is not a separate skill — it is the question "find $x$ when $f(x) = 0$", which is why the zero and the $x$-intercept turn up in every window you choose.

Both directions in a context

A rideshare charges $C(m) = 2.5m + 3.5$ dollars for a ride of $m$ miles.

• Given the input. A $12$-mile ride: $C(12) = 2.5(12) + 3.5 = 33.5$, so the fare is $\$33.50$.
• Given the output. A fare of $\$28.50$: solve $2.5m + 3.5 = 28.5$, giving $2.5m = 25$ and $m = 10$, so the ride was $10$ miles.

The second is the question a rider actually asks, and it is the one that requires solving rather than substituting. Notice that both answers carry units, and that the units differ — dollars for the output, miles for the input.

Worked examples

Example 1 — Evaluating, including a negative input

For $f(x) = -3x + 8$, find $f(2)$ and $f(-4)$.

$f(2) = -3(2) + 8 = 2$. $f(-4) = -3(-4) + 8 = 12 + 8 = 20$.

Answer: $f(2) = 2$ and $f(-4) = 20$

Example 2 — Recovering the input

For $f(x) = -3x + 8$, find $x$ when $f(x) = 23$.

$-3x + 8 = 23$, so $-3x = 15$ and $x = -5$.

Answer: $x = -5$

Example 3 — Reading a graph in both directions

Using the graph of $f(x) = 2x - 3$, find $f(1)$ and find $x$ when $f(x) = 1$.

Up from $1$ meets the line at height $-1$. Over from $1$ on the $y$-axis meets the line above $x = 2$.

Answer: $f(1) = -1$; and $f(x) = 1$ at $x = 2$. Both check algebraically: $2(1) - 3 = -1$ and $2(2) - 3 = 1$.

Example 4 — A non-integer answer

For $g(x) = 4x + 1$, find $x$ when $g(x) = 3$.

$4x + 1 = 3$, so $4x = 2$ and $x = \tfrac12$.

Answer: $x = \tfrac12$. An input recovered this way need not be a whole number.

Example 5 — Both directions in context

A room cools according to $T(h) = 68 - 4h$ degrees after $h$ hours. Find $T(6)$, and find when the room reaches $20$ degrees.

$T(6) = 68 - 24 = 44$. And $68 - 4h = 20$ gives $4h = 48$, so $h = 12$.

Answer: After $6$ hours the room is $44$ degrees. It reaches $20$ degrees after $12$ hours.

Guided practice

61. Use the figure showing the path up from $x = 4$. What is $f(4)$, and describe the two travels the dashed path makes.
62. Confirm that value algebraically from $f(x) = 2x - 3$.
63. Using the same rule, find $f(-1)$ algebraically.
64. Use the figure showing the path from $y = -7$. What input produced that output, and describe the path.
65. Confirm that input algebraically by solving an equation.
66. Explain in one or two sentences how the questions in items 61 and 64 differ, and what the two answers have in common.

Independent practice

67. For $f(x) = -3x + 8$, find each value. a) $f(0)$ b) $f(2)$ c) $f(-4)$ d) $f\!\left(\tfrac13\right)$
68. For $f(x) = \tfrac12 x - 6$, find each value. a) $f(10)$ b) $f(-8)$ c) $f(0)$ d) $f(3)$
69. For $f(x) = -3x + 8$, find $x$ in each case. a) $f(x) = -7$ b) $f(x) = 8$ c) $f(x) = 23$ d) $f(x) = 0$
70. For $g(x) = 4x + 1$, find $x$ in each case. a) $g(x) = 13$ b) $g(x) = -11$ c) $g(x) = 1$ d) $g(x) = 3$
71. Use the graph of $f(x) = 2x - 3$ in the figure for reading $f(x)$. Read $f(0)$, $f(1)$, and $f(2)$ off the graph, then confirm all three algebraically.
72. Use that same graph. Find the $x$ for which $f(x) = 1$, and the $x$ for which $f(x) = -3$. Say which axis you started from.
73. Application. A rideshare charges $C(m) = 2.5m + 3.5$ dollars for $m$ miles. Find $C(12)$, and find the mileage of a ride that cost $\$28.50$. Write one sentence for each answer, with units.
74. Application. A room cools according to $T(h) = 68 - 4h$ degrees after $h$ hours. Find $T(6)$, and find when the room reaches $20$ degrees. Give units for both.
75. Error analysis. Asked to find $x$ when $f(x) = 10$ for $f(x) = 2x + 4$, a student computes $f(10) = 24$ and answers $24$. Identify the error, say which question the student answered, and give the correct $x$.
76. Reasoning. Explain why recovering $x$ from a given $f(x)$ is always the work of solving an equation, while finding $f(x)$ from a given $x$ never is.
77. Technology. For $f(x) = 0.75x - 2.5$, find $f(8)$ algebraically, then describe how the table feature of a graphing calculator would confirm it.
78. Application. Use the bike-rental graph of $C(h) = 5h + 20$ from Lesson 7.2. Find $C(5)$ and find the number of hours that costs $\$55$, reading each off the graph and confirming it algebraically.

Exit ticket 7.4

79. For $f(x) = -2x + 9$, find $f(-3)$, and find $x$ when $f(x) = 1$.
80. Use the figure for reading $f(x)$ off a graph. State $f(4)$, and state the $x$ for which $f(x) = 5$. Explain why one marked point answers both.
81. Application. A savings balance falls according to $B(w) = 250 - 18w$ dollars after $w$ weeks. Find $B(9)$, and find the week the balance reaches $\$34$.
82. Describe, in words, the two dashed paths in this lesson's figures: which axis each starts from, and which axis each ends at.

Lesson 7.5 — Comparing the Four Representations

The same function, four ways

A.F.1h names four representations — algebraic, graphical, tabular, contextual — and asks you to compare and contrast the characteristics of linear functions across them. This is the capstone of the linear sequence, because until now you have been working inside one representation at a time.

The figure carries one relationship in all four forms at once.

Contextual. A pool holds $400$ gallons and drains at $25$ gallons per minute; $W$ is the gallons left after $t$ minutes.

Algebraic. $W(t) = 400 - 25t$, with slope $-25$, $W$-intercept $(0,400)$, and zero $t = 16$.

Tabular.

Graphical. A segment falling from $(0,400)$ to $(16,0)$.

All four describe the same pool. None of them contains information the others lack. But they do not display the same things with the same effort, and that is what the comparison is about.

Where each characteristic is cheapest to find

Three consequences are worth stating outright.

• The equation is the only exact one. A graph read by eye gives $f(3) \approx 2.7$; the equation gives $\tfrac{27}{10}$. When a question says "determine," reach for the equation.
• The graph is the only one you can take in at a glance. Nobody sees a trend in an equation the way they see it in a falling line.
• The table shows constancy. Equal steps of $4$ minutes drop $W$ by the same $100$ gallons every time. That repeated difference is linearity, and it is visible in the table and nowhere else so plainly.

And one warning. A table does not always display the $y$-intercept. Given

the slope is $3$, but the $y$-intercept is not $7$ — the table simply does not contain the input $0$. Stepping back one row from $(1,7)$ gives $(0,4)$, so the function is $y = 3x + 4$.

Comparing two different functions

The harder version of bullet h is comparing two linear functions that arrive in different representations. The method is to convert both to whatever form answers the question, and slope-intercept form usually does.

Two courts rent by the hour. Plan A charges a $\$60$ membership plus $\$10$ per hour. Plan B charges $\$25$ per hour and nothing else.

$$A(x) = 10x + 60 \qquad B(x) = 25x$$

• Rate of change. $B$ climbs faster: $\$25$ per hour against $\$10$ per hour.
• Initial value. $A$ starts higher: $\$60$ against $\$0$.
• Where they agree. Set the two outputs equal — a one-variable equation of the kind Chapter 2 solved:

$$10x + 60 = 25x \quad\Rightarrow\quad 60 = 15x \quad\Rightarrow\quad x = 4$$

and $A(4) = B(4) = 100$. The graph marks $(4, 100)$, and the two computations agree.

• Which is better, and when. Below $4$ hours, $B$ is cheaper — at $2$ hours, $B$ costs $\$50$ and $A$ costs $\$80$. Above $4$ hours, $A$ is cheaper — at $6$ hours, $A$ costs $\$120$ and $B$ costs $\$150$. "Which plan is better" has no answer without an amount of use, and saying so is part of the comparison.

When the slopes are equal, there is no crossing at all. $f(x) = 4x + 20$ and a table stepping up $8$ for every $2$ across starting at $30$ both have slope $4$; their graphs are parallel, and the one that starts lower stays lower forever. That is Chapter 6's parallel-lines fact, seen from the comparison side.

Choosing a representation on purpose

Given a choice, pick the one that answers the question asked.

• "Exactly how much for $17$ hours?" — the equation. Substitute.
• "When does Plan A overtake Plan B?" — the graph. Look at the crossing.
• "Is this even linear?" — the table. Check for a constant difference over equal steps.
• "What does the slope mean?" — the context. Only the story supplies units.

Worked examples

Example 1 — Building all four

A lawn service charges a $\$15$ trip fee plus $\$6$ per hour. Give the algebraic, tabular, and graphical forms.

Answer: Algebraic: $f(x) = 6x + 15$. Tabular: at $x = 0, 1, 2, 3$ the outputs are $15, 21, 27, 33$. Graphical: a line from $(0,15)$ rising $6$ dollars per hour, best drawn at $1$ hour and $5$ dollars per cell.

Example 2 — A table without the intercept

A table lists $(1,7)$, $(2,10)$, $(3,13)$. Give the slope and the $y$-intercept.

Each step of $1$ in $x$ raises $y$ by $3$, so $m = 3$. Stepping back from $(1,7)$ gives $(0,4)$.

Answer: Slope $3$; $y$-intercept $(0,4)$; the function is $y = 3x + 4$.

Example 3 — Comparing across two representations

$f(x) = 4x + 20$ is given by an equation; $g$ is given by the table $(0,30)$, $(2,38)$, $(4,46)$, $(6,54)$. Compare them.

$g$ rises $8$ for every $2$, so its slope is $4$, and its table shows the input $0$ with output $30$.

Answer: Same rate of change, $4$. $g$ has the larger initial value, $30$ against $20$. Their graphs are parallel, so they never meet, and $g$ is greater than $f$ by $10$ at every input.

Example 4 — Comparing with a crossing

Gym A charges $\$25$ a month plus $\$5$ per class. Gym B is given by the table $(0,10)$, $(2,30)$, $(4,50)$, $(6,70)$. Which is cheaper for $3$ classes, and when do they cost the same?

$A(c) = 5c + 25$. Gym B rises $20$ per $2$ classes, so $B(c) = 10c + 10$.

At $c = 3$: $A(3) = 40$ and $B(3) = 40$ — the same. Solving $5c + 25 = 10c + 10$ gives $15 = 5c$, so $c = 3$.

Answer: They cost the same, $\$40$, at exactly $3$ classes. Below $3$ classes B is cheaper; above $3$, A is cheaper because its rate is lower.

Example 5 — Which representation to hand over

Someone wants to know the exact cost of a $17$-hour rental under $C(h) = 5h + 20$. Which representation, and why?

Answer: The equation. Substituting gives $C(17) = 105$ exactly, while reading a graph at $17$ gives an estimate and a table that stops at $8$ gives nothing at all.

Guided practice

83. Use the four-representations figure. Write the algebraic rule, and state the slope, the intercept, and the zero, each with its meaning in the story.
84. In that same figure, what does the table make visible that the equation does not display at a glance?
85. In that same figure, what does the graph show faster than any of the other three?
86. Find $W(10)$ for that function. Say which representation you used and why the others were worse choices.
87. Use the two-plans figure. Which plan is cheaper at $2$ hours, and which at $6$ hours? Give both costs each time.
88. In that same figure, confirm the marked crossing point by solving one equation in one variable.

Independent practice

89. A lawn service charges a $\$15$ trip fee plus $\$6$ per hour. Write the algebraic rule, build a table for $x = 0, 1, 2, 3$, describe the graph, and write the contextual sentence that names the slope with units.
90. $f(x) = 4x + 20$ is given as an equation; $g$ is given by the table $(0,30)$, $(2,38)$, $(4,46)$, $(6,54)$. Compare their rates of change and their initial values, and say whether the two graphs ever meet.
91. $f(x) = -5x + 40$ is given as an equation; $g$ is described as a line falling from $(0,25)$ to $(5,0)$. Compare the two slopes and the two initial values, and say what the comparison tells you about the graphs.
92. Three functions arrive in three forms: the equation $y = 3x - 1$; a table whose output rises by $4$ for every $1$ the input rises; and a context in which a seller earns $\$2.50$ per item. Rank them by rate of change, and say what made each rate visible.
93. Application. Gym A charges $\$25$ a month plus $\$5$ per class. Gym B is given by the table $(0,10)$, $(2,30)$, $(4,50)$, $(6,70)$. Find the cost of $3$ classes at each gym, find where the two agree, and say which gym is better for someone taking $6$ classes a month.
94. Reasoning. Name the representation you would use to find a zero exactly, and the representation you would use to see a trend instantly. Justify each choice in one sentence.
95. Error analysis. Given the table $(1,7)$, $(2,10)$, $(3,13)$, a student reports a $y$-intercept of $(0,7)$. Identify the error and give the correct slope, intercept, and equation.
96. Application. A scooter rental charges $\$4$ to unlock plus $\$0.25$ per minute. Write the equation, build a table at $t = 0, 10, 20, 30$ minutes, describe the graph including a scale for each axis, and say what the slope means with units.
97. Use the two-plans figure. Which plan's equation has the larger $y$-intercept, what does that number mean in the story, and why does the plan with the larger $y$-intercept still win in the long run?
98. Reasoning. A table says $f(3) = 11$ but the equation $f(x) = 3x + 1$ gives $10$. Explain what you should conclude and what you should do next.