Write and match linear functions

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Difficulty: Beginner
Estimated time: 30 minutes
Domains: Algebra
Techniques: Slope-intercept-formTwo-independent-factsPoint-substitutionRepresentation-matching

What you’ll learn

Tell when different information describes the same line.
Write a linear function from a slope and a point, two points, or a table.
Match an equation to a graph, a table or another equation.
Decide when to build the rule by hand and when to graph the answer choices, then check your answer against two independent facts.

Prerequisites

Why this matters on the SAT

Build the function you need

The SAT can hand you a line in lots of disguises: a table, two points, a graph, or an equation that isn't solved for $y$ yet. Underneath, the job is the same every time. Find the line's slope and one point it passes through, then pick the function that has both.

Solution to the example

Start with the first two rows. The input goes up by $3$ , from $-2$ to $1$ , and the output goes down by $6$ , from $9$ to $3$ . So the slope is

m=\frac{-6}{3}=-2.

Now find $b$ . Put the point $(1,3)$ into $f(x)=mx+b$ :

3=-2(1)+b,

so $b=5$ . The function is

f(x)=-2x+5.

That's choice A. Look at what you used: one slope and one point. Those two facts are all it takes to pin down a line.

SAT example

Some values of the linear function $f$ are shown in the table.

$x$	$f(x)$
$-2$	$9$
$1$	$3$
$4$	$-3$

Which equation defines $f$ ?

A
$f(x)=-2x+5$
B
$f(x)=2x+5$
C
$f(x)=-2x+3$
D
$f(x)=-\frac12x+5$

Two facts determine one line

A linear function is usually written in slope-intercept form:

f(x)=mx+b.

Here $x$ is the input and $f(x)$ is the output. The number $m$ is the slope, how much the output changes each time $x$ goes up by $1$ . The number $b$ is the output when $x=0$ , so the graph crosses the $y$ -axis at $(0,b)$ .

If a question uses $y$ instead of $f(x)$ , nothing changes:

y=mx+b \qquad\text{and}\qquad f(x)=mx+b

describe the same link between input and output.

You already know how to find both pieces. Slope and rate of change tells you how the output changes. Intercepts and starting values tell you where the line sits when the input is $0$ . Put those two facts together and you get exactly one line, as long as it isn't vertical.

The two facts have to be independent, which means the second one tells you something new. The point $(0,5)$ and "the $y$ -intercept is $5$ " say the same thing, so they count as one fact, not two. A slope plus a point works. So do two points, as long as their $x$ -values are different.

One point on its own isn't enough, because lots of lines pass through the same point. Here's the phrase to remember: one point, many lines; two facts, one line.

Check your understanding:

The graph of a linear function passes through $(2,5)$ . Why isn't that enough to write its rule? What extra fact would be enough?

Don't forget flat lines. A horizontal line is still a linear function. If two different inputs give the same output, the slope is $m=0$ and the rule is a constant, like $f(x)=17$ . A vertical line is different. One input would have more than one output, so it can't be written as $f(x)$ .

Start from the information you have

Aim for $f(x)=mx+b$ . Look at what the question gives you, and work out only the piece that's missing.

Write the function by hand

If you have the slope and the $y$ -intercept, say $m=4$ and $b=-1$ , write $f(x)=4x-1$ straight away.
If you have the slope and some other point, put the point into $f(x)=mx+b$ and solve for $b$ .
If you have two points or a table, find $m$ from two rows, then use either row to find $b$ .
If you have an equation like $3x+2y=14$ , get $y$ by itself so you can read $m$ and $b$ .
If you have a graph, read two labeled points, or the intercept and the slope, and check the scale on each axis.

Match the choices without building the rule

Put a given point into a choice, and cross the choice out if the two sides don't match.
A line that falls from left to right needs a negative slope, and one that rises needs a positive slope.
If you can see the $y$ -intercept, cross out any choice with a different value at $x=0$ .
If the choices are complete equations and the data are awkward, like $(1.6,8.7)$ , plot the points in Desmos, graph the choices and keep the line that passes through both points.

Core rule: a choice has to match a slope and a point, or two points. One matching point can leave several choices standing.

With clean numbers, building the rule by hand is usually quickest, so start there. Graph the choices only when typing them in is less work than finding the rule. If a question gives several measured pairs or asks for a line of best fit, use Desmos regression from a supplied table instead.

From a slope and a point

Say a linear function $g$ has slope $-\frac12$ and passes through $(6,4)$ . You already know $m$ , so start with

g(x)=-\frac12x+b.

The point $(6,4)$ means input $6$ gives output $4$ . Put $6$ in for $x$ and $4$ in for $g(x)$ :

\begin{aligned} 4&=-\frac12(6)+b\\[1.4em] 4&=-3+b\\[1.4em] b&=7. \end{aligned}

g(x)=-\frac12x+7.

Here's the trap. The point's output, $4$ , is not the $y$ -intercept. If you wrote $g(x)=-\frac12x+4$ , you'd get $g(6)=-3+4=1$ , not $4$ . Only a point with input $0$ hands you $b$ directly. For any other point, you solve for $b$ .

Try it yourself:

A linear function has slope $-3$ and passes through $(2,1)$ . Start with $f(x)=-3x+b$ , put in the point, and find $b$ .

Check your understanding:

What's $b$ for the function in Try it yourself, and what's the full rule?

From two points or a table

Each row of a table is a point $(x,f(x))$ on the graph. So a table works just like two points: pick two rows, find the slope, then put either row in to find $b$ .

When you find the slope, subtract in the same order on the top and the bottom:

m=\frac{f(x_2)-f(x_1)}{x_2-x_1}.

If you start with the second point on top, start with it on the bottom too. And watch the first row of a table: its output is $b$ only if its input is $0$ .

From an equation or graph

If an equation doesn't have $y$ by itself yet, first isolate the output variable. For example,

3x+2y=14

becomes

\begin{aligned} 2y&=-3x+14\\[1.4em] y&=-\frac32x+7. \end{aligned}

Now you can read the slope, $-\frac32$ , and the $y$ -intercept, $7$ .

You may also see a line in point-slope form:

y-y_1=m(x-x_1).

It shows a slope $m$ and one point $(x_1,y_1)$ on the line. There's no new method to learn. Either expand it into slope-intercept form, or check that the point makes both sides equal $0$ and that the slope is right. Watch the signs: $y-5=2(x+1)$ passes through $(-1,5)$ , not $(1,5)$ , because $x+1$ is $x-(-1)$ .

On a graph, trust the labeled numbers, not how steep the line looks. The two axes can use different scales, and that can make the same slope look steeper or flatter. Read two points you're sure of, then find the rule the same way you would from two points.

Check your understanding:

Rewrite $4x-2y=10$ as $y=mx+b$ . What slope and $y$ -intercept does it show?

Common mistake:

A common slip is picking a choice as soon as it passes through one given point. Plenty of wrong lines go through that point too. Before you commit, check the slope as well, or put the second given point into your answer.

Example: Write the rule from two points

Worked example

The graph of the linear function $f$ passes through the points $(2,7)$ and $(6,-1)$ . Which equation defines $f$ ?

A
$f(x)=-2x+11$
B
$f(x)=2x+3$
C
$f(x)=-2x+7$
D
$f(x)=-\frac12x+8$

Step 1

Find the slope

Divide the change in output by the change in input, taking the points in the same order:

\begin{aligned} m &=\frac{-1-7}{6-2}\\[1.4em] &=\frac{-8}{4}\\[1.4em] &=-2. \end{aligned}

A negative slope makes sense here: as $x$ goes up from $2$ to $6$ , the output drops from $7$ to $-1$ .

Step 2

Find the intercept

Start from $f(x)=mx+b$ with $m=-2$ , and put in $(2,7)$ :

\begin{aligned} 7&=-2(2)+b\\[1.4em] 7&=-4+b\\[1.4em] b&=11. \end{aligned}

So the function is

f(x)=-2x+11.

Step 3

Match and check

Choice A has the slope and intercept you found. Now check the point you haven't used yet:

f(6)=-2(6)+11=-1.

That matches $(6,-1)$ , so the answer is A.

Calculator loads as you approach

Both marked points sit on the line. Change only the slope, or only the constant, and watch the line miss the points.

Check your understanding:

Choices B and D both pass through $(2,7)$ . Why is each one still wrong?

Practice problems

Your turn. Each problem starts from different information, so decide first which two facts you have.

Write a function from a slope and a point

Practice problem

A linear function $g$ has slope $\frac25$ and passes through the point $(5,1)$ . Which equation defines $g$ ?

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Solve this one by hand. To check, plot

(5,1)

and graph your rule to see the line pass through it.

Match a graph to point-slope form

Practice problem

The graph of the linear function $f$ is shown. Which equation defines $f$ ?

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Reading the marked points by hand is quickest. To check, plot both points and graph your choice to see the line pass through them.

Graph the answer choices

Practice problem

The graph of the linear function $h$ passes through $(1.6,8.7)$ and $(6.4,-3.9)$ . Which equation defines $h$ ?

Calculator loads as you approach

The decimals are messy and the choices are complete equations, so graphing the choices is the quicker way here.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Turn whatever you're given into two independent facts about the line.
Use $f(x)=mx+b$ : find the slope first, then put in a point to find $b$ . A point's output is $b$ only when its input is $0$ .
Table rows are points. In an equation, get the output variable by itself. On a graph, read labeled coordinates and check the scales.
Point-slope form shows a slope and one point, and it's the same line as slope-intercept form.
With clean numbers, build the rule by hand. With awkward data and complete choices, graph the points and the choices when that's less work.
Before you choose, check a slope and a point, or two points. One matching point isn't enough.

Next lesson

Find inputs and outputs of linear functions

Use a linear function to evaluate an output or recover the input that produces a given output.

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1,570 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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