Linear Functions Pattern - Abstract Functions

This pattern tests your ability to work with linear functions abstractly — finding unknown constants, evaluating combinations of functions, and reasoning about function definitions that use parameters instead of numbers.

 

Find an Unknown Constant by Substitution

The most common type: you're given a function with an unknown constant and one function value. Plug in the given input, set equal to the given output, and solve for the constant.

For the linear function $f$, $f(x) = mx + 5$, where $m$ is a constant. If $f(3) = 26$, what is the value of $m$?

A) $-7$
B) $3$
C) $5$
D) $7$

Substitute $x = 3$ into $f(x) = mx + 5$ and set it equal to $26$:

$$m(3) + 5 = 26$$

$$3m = 21$$

$$m = 7$$

The answer is D.

When the given input is $0$, the problem simplifies since the variable term vanishes — you're left with just the constant.

$p(y) = m - 2y$

For the function $p$, the constant $m$ is the $y$-intercept of its graph. If $p(0) = -8$, what is the value of $m$?

A) $-8$
B) $-2$
C) $4$
D) $8$

Substitute $y = 0$: $p(0) = m - 2(0) = m$. Since $p(0) = -8$, we get $m = -8$. The answer is A. When the input is $0$, the function value equals the constant term directly.

Watch for negative inputs — the double negative is where most errors happen.

The linear function $g$ is defined by $g(x) = 6x + a$, where $a$ is a constant. If $g(-2) = 20$, what is the value of $a$?

A) $-2$
B) $6$
C) $8$
D) $32$

Substitute $x = -2$: $6(-2) + a = 20$, so $-12 + a = 20$, giving $a = 32$. The answer is D. A common mistake is computing $6(-2) = 12$ instead of $-12$.

 

Find the Constant, Then Evaluate at a New Input

A two-step version: use the given value to find the unknown constant, then evaluate the function at a different input.

The linear function $h$ is defined by $h(x) = kx - 25$, where $k$ is a constant. If $h(-4) = 15$, what is the value of $h(6)$?

Step 1: Find $k$. Substitute $x = -4$:

$$k(-4) - 25 = 15$$

$$-4k = 40$$

$$k = -10$$

Step 2: Evaluate $h(6)$. Now that $h(x) = -10x - 25$:

$$h(6) = -10(6) - 25 = -60 - 25 = -85$$

The answer is $-85$.

This also appears with more complex function forms where the constant multiplies an expression.

A function $h$ is defined as $h(t) = a(t + 5) + 3t$, where $a$ is a constant. If $h(-2) = 12$, what is the value of $h(4)$?

Step 1: Find $a$. Substitute $t = -2$:

$$a(-2 + 5) + 3(-2) = 12$$

$$3a - 6 = 12$$

$$a = 6$$

Step 2: Evaluate $h(4)$. Now $h(t) = 6(t + 5) + 3t$:

$$h(4) = 6(4 + 5) + 3(4) = 6(9) + 12 = 54 + 12 = 66$$

The answer is $66$.

 

Evaluate a Linear Combination of Two Functions

You're given two function definitions and asked to evaluate an expression that mixes them — like $3p(4) + q(4)$ or $2a(-1) - 5b(-1)$.

If $p(t) = t - 5$ and $q(t) = 3t + 1$, what is the value of $3p(4) + q(4)$?

A) $20$
B) $12$
C) $38$
D) $10$

Evaluate each function at $t = 4$:

$p(4) = 4 - 5 = -1$

$q(4) = 3(4) + 1 = 13$

Now compute: $3p(4) + q(4) = 3(-1) + 13 = -3 + 13 = 10$. The answer is D.

With negative inputs, keep careful track of signs through each substitution.

The functions $a$ and $b$ are defined such that $a(n) = 5n - 9$ and $b(n) = n + 1$. What is the value of $2a(-1) - 5b(-1)$?

A) $-24$
B) $-33$
C) $-14$
D) $-28$

Evaluate each: $a(-1) = 5(-1) - 9 = -14$ and $b(-1) = -1 + 1 = 0$.

Compute: $2(-14) - 5(0) = -28 - 0 = -28$. The answer is D.

 

Find $a + b$ from Intercepts

Given a function, find the x-intercept $(a, 0)$ and y-intercept $(0, b)$, then compute $a + b$.

The function $k$ is defined by $k(t) = -6t - 42$. The graph of $y = k(t)$ in the ty-plane has a t-intercept at $(a, 0)$ and a y-intercept at $(0, b)$, where $a$ and $b$ are constants. What is the value of $a + b$?

A) $-35$
B) $-42$
C) $-48$
D) $-49$

Y-intercept: Set $t = 0$: $k(0) = -6(0) - 42 = -42$. So $b = -42$.

T-intercept: Set $k(t) = 0$: $-6t - 42 = 0 \Rightarrow t = -7$. So $a = -7$.

$a + b = -7 + (-42) = -49$. The answer is D.

The zero of a function is just another name for the x-intercept. Set the function equal to $0$ and solve.

The linear function $h$ is defined by $h(x) = \dfrac{2}{3}x - 12$. What is the zero of the function?

Set $h(x) = 0$: $\dfrac{2}{3}x - 12 = 0$, so $\dfrac{2}{3}x = 12$, giving $x = 18$. The answer is $18$.

 

Combining Functions to Form a New Function

You may be given two functions and told that a third is their sum or difference. Usually you need to simplify and find an intercept.

Two linear functions are given by $j(x) = 5x - 8$ and $k(x) = 3x + 12$. A third function, $m$, is defined by the difference $m(x) = j(x) - k(x)$. What is the $x$-coordinate of the point where the graph of $y = m(x)$ crosses the $x$-axis?

Combine: $m(x) = (5x - 8) - (3x + 12) = 5x - 8 - 3x - 12 = 2x - 20$.

Set $m(x) = 0$: $2x - 20 = 0 \Rightarrow x = 10$. The answer is $10$.

When the functions involve fractions, just add the coefficients carefully.

The functions $p$ and $q$ are defined as $p(x) = \dfrac{2}{5}x + 10$ and $q(x) = \dfrac{3}{5}x - 20$. If $r(x) = p(x) + q(x)$, what is the $x$-coordinate of the $x$-intercept of the graph of $y = r(x)$?

Combine: $r(x) = \left(\dfrac{2}{5}x + 10\right) + \left(\dfrac{3}{5}x - 20\right) = \dfrac{2}{5}x + \dfrac{3}{5}x + 10 - 20 = x - 10$.

Set $r(x) = 0$: $x - 10 = 0 \Rightarrow x = 10$. The answer is $10$.

 

Table-Based Function Transformation

You may see a table of values for one function, then be asked about a transformed version like $j(n) = m(n) - 18$.

A table of values for the linear function $m$ is shown.

$n$	$m(n)$
$0.5$	$-7$
$1$	$-2$
$2$	$8$

If the function $j$ is defined by $j(n) = m(n) - 18$, which equation defines $j$?

A) $j(n) = 10n + 10$
B) $j(n) = 10n - 12$
C) $j(n) = 10n + 6$
D) $j(n) = 10n - 30$

Step 1: Find $m(n)$. Use the table to get the slope: $\dfrac{-2 - (-7)}{1 - 0.5} = \dfrac{5}{0.5} = 10$. Using point $(1, -2)$: $m(n) = 10n + b$, so $-2 = 10(1) + b$, giving $b = -12$. Thus $m(n) = 10n - 12$.

Step 2: Find $j(n)$. $j(n) = m(n) - 18 = (10n - 12) - 18 = 10n - 30$. The answer is D.

 

Substituting an Expression as Input

In harder problems, the function is evaluated at an expression like $k - 5$ or $q + 3$ instead of a number. Treat the expression as the input and simplify.

The linear function $h$ is defined by $h(x) = mx + 12$, where $m$ is a constant. If $h(k - 5) = \dfrac{k}{3}$, where $k$ is a constant, which expression represents the value of $m$?

A) $\dfrac{k + 36}{3(k - 5)}$
B) $\dfrac{k}{3(k - 5)}$
C) $\dfrac{k - 36}{3(k - 5)}$
D) $\dfrac{k - 36}{3k - 5}$

Replace $x$ with $k - 5$ in the function:

$$m(k - 5) + 12 = \dfrac{k}{3}$$

$$m(k - 5) = \dfrac{k}{3} - 12 = \dfrac{k - 36}{3}$$

$$m = \dfrac{k - 36}{3(k - 5)}$$

The answer is C. The key step is converting $12$ to $\dfrac{36}{3}$ to combine fractions.

 

Functions Defined with Parametric Inputs: $g(ax) = x - 1$

The trickiest type: a function is defined as $g(ax) = x - 1$ for all $x$, with $a$ as an unknown constant. You're given something like $g(5) = 9$ and need to find $a$. The strategy is to find what value of $x$ makes $ax = 5$.

The function $g$ is defined by $g(ax) = x - 1$ for all $x$, where $a$ is a nonzero constant. If $g(5) = 9$, what is the value of $a$?

We need $ax = 5$, so $x = \dfrac{5}{a}$. Substituting into $g(ax) = x - 1$:

$$g(5) = \dfrac{5}{a} - 1$$

Since $g(5) = 9$:

$$\dfrac{5}{a} - 1 = 9$$

$$\dfrac{5}{a} = 10$$

$$a = \dfrac{1}{2}$$

The answer is $\dfrac{1}{2}$.