Linear or Nonlinear? Comparing Patterns of Change | Grade 8 Math | LevelKite

Some patterns change by the same amount at every step. Others speed up or slow down. This lesson gives you a reliable way to tell which kind you are looking at.

First, recall that a function pairs each permitted input with exactly one output. The set of permitted inputs is called the domain. Different inputs may share an output, but one input can never have two outputs.

You already know one family of functions: linear functions, whose rules can be written as $y = mx + b$ with fixed numbers $m$ and $b$ . Functions outside that family are called nonlinear. You will learn to tell the two kinds apart, give your own examples of nonlinear functions, and compare how the two kinds change.

Start with two rules. For both, the domain is all real numbers $x$ , and $y$ stands for the output.

Rule 1: $y = 2x + 1$ . Multiply the input by $2$ , then add $1$ .
Rule 2: $y = x^2$ . The raised $2$ tells you to multiply the input by itself, so $x^2 = x \times x$ . This is called squaring the input. For example, $3^2 = 3 \times 3 = 9$ .

Each rule gives exactly one output for every input, so both are functions. In Rule 2, two different inputs can share an output: $(-2) \times (-2) = 4$ , because a negative times a negative is positive, and $2 \times 2 = 4$ . That is allowed in a function.

The table shows both rules for the inputs $0$ , $1$ , $2$ and $3$ . Each change column shows how much that rule's output rises from the row above.

Rule 1 and Rule 2 for the inputs 0, 1, 2 and 3

Input x	Rule 1 output	Change in Rule 1	Rule 2 output	Change in Rule 2
$0$	$2(0) + 1 = 1$	no row above	$0 \times 0 = 0$	no row above
$1$	$2(1) + 1 = 3$	$3 - 1 = 2$	$1 \times 1 = 1$	$1 - 0 = 1$
$2$	$2(2) + 1 = 5$	$5 - 3 = 2$	$2 \times 2 = 4$	$4 - 1 = 3$
$3$	$2(3) + 1 = 7$	$7 - 5 = 2$	$3 \times 3 = 9$	$9 - 4 = 5$

Read down the two change columns. From each row to the next, $x$ goes up by the same step, $1$ .

Rule 1's output rises by $2$ every time. That is a constant difference.
Rule 2's output rises by $1$ , then $3$ , then $5$ . Those are changing differences.

Rule 1's constant difference is no accident. Raise any input $x$ by $1$ . By the distributive property, $2(x + 1) + 1 = 2x + 2 + 1 = (2x + 1) + 2$ . The new output is the old output plus $2$ , whatever $x$ is. Every rule $y = mx + b$ works this way: $m(x + 1) + b = mx + m + b = (mx + b) + m$ , so a step of $1$ in $x$ always changes $y$ by $m$ .

Rule 2 has no such fixed amount. In the table, each step multiplies a larger number by itself, and the three changes are all different.

Now look at the same two rules on a coordinate plane. In the figure below, the horizontal axis shows the input $x$ and the vertical axis shows the output $y$ . Each pair is written as (input, output).

Line A is the graph of Rule 1, $y = 2x + 1$ . Rule 1's table pairs $(0, 1)$ , $(1, 3)$ , $(2, 5)$ and $(3, 7)$ all lie on it.

The four labeled points are computed from Rule 2, $y = x^2$ . They are samples of its graph: Rule 2 has an output for every real input, but only these four exact points are plotted. No curve is drawn through them.

Line A is the graph of

y = 2x + 1

. The four labeled points are samples of

y = x^2

at

x = 0, 1, 2, 3

.

Are the four samples in the figure above collinear, meaning all on one straight line? Decide from the coordinates, not from how the picture looks. The slope between two points is the change in $y$ divided by the change in $x$ . It is also called the rate of change between them.

On Line A, any two points give the slope $2$ . For example, from $(0, 1)$ to $(3, 7)$ it is $\frac{7 - 1}{3 - 0} = \frac{6}{3} = 2$ .
For the samples, the slope from $(0, 0)$ to $(1, 1)$ is $\frac{1 - 0}{1 - 0} = 1$ , but the slope from $(1, 1)$ to $(2, 4)$ is $\frac{4 - 1}{2 - 1} = 3$ .

A line through $(0, 0)$ and $(1, 1)$ rises $1$ for each step of $1$ , so it would reach $(2, 2)$ . The sample is $(2, 4)$ instead. Because $1 \neq 3$ , the samples are not collinear, so Rule 2 is not linear.

Key concept

Linear means one constant rate of change

A function is linear when one rule $y = mx + b$ , with fixed numbers $m$ and $b$ , gives all of its input-output pairs. Then a step of $h$ in $x$ always changes $y$ by $mh$ , because $m(x + h) + b = (mx + b) + mh$ . So the slope between any two of its points is $\frac{mh}{h} = m$ . Here $h \neq 0$ , because two different points of a function have different inputs. On a graph, all of its points lie on one straight line that is not vertical.

A function is nonlinear when no such rule fits all of its pairs. To prove it, find three of its points that give two unequal slopes. No single line can pass through all three.

\frac{y_2 - y_1}{x_2 - x_1} = m \quad \text{for every pair of points with } x_1 \neq x_2

Two cautions about straight lines.

Linear does not mean proportional. A proportional relationship has the form $y = kx$ , where $k$ is a fixed number: its graph passes through the origin $(0, 0)$ , and $\frac{y}{x} = k$ for every $x \neq 0$ . Rule 1 is linear but not proportional, because its $b$ is $1$ , not $0$ . Its graph passes through $(0, 1)$ instead of the origin, and $\frac{y}{x}$ is not constant: it is $\frac{3}{1} = 3$ at $x = 1$ but $\frac{5}{2} = 2.5$ at $x = 2$ .
A vertical line, such as $x = 3$ , is straight, but it is not the graph of a function of $x$ : it pairs the one input $3$ with many outputs. Its slope is undefined, because the change in $x$ between any two of its points is $0$ and you cannot divide by $0$ .

Worked example

Linear or nonlinear?

Decide whether each function is linear or nonlinear. (a) $y = 7 - 3x$ for all real numbers $x$ . (b) A function whose table shows the input-output pairs $(0, 3)$ , $(2, 7)$ and $(4, 15)$ .

For (a), try to write the rule in the form y = mx + b. Subtracting 3x is the same as adding negative 3x, and changing the order of two terms does not change their sum.
$y = 7 - 3x = 7 + (-3x) = -3x + 7$
The rule now has the form y = mx + b, so read off m and b. Rule (a) is linear for every real input, and its output falls by 3 each time x rises by 1.
$m = -3, \quad b = 7$
For (b) there is no rule, so use the points. Find the slope between (0, 3) and (2, 7). The inputs are different, so the change in x is not 0 and the division is allowed.
$\frac{7 - 3}{2 - 0} = \frac{4}{2} = 2$
Find the slope between (2, 7) and (4, 15) in the same way.
$\frac{15 - 7}{4 - 2} = \frac{8}{2} = 4$
Compare the two slopes. A linear function gives the same slope between every pair of its points. These slopes are unequal, so no single line passes through all three points, whatever other pairs the function may have.
$2 \neq 4$

Answer

Function (a) is linear, with $m = -3$ and $b = 7$ . Function (b) is nonlinear, because its three points give the unequal slopes $2$ and $4$ .

The four table pairs have unequal gaps in

x

, yet all four lie on Line A,

y = 2x + 1

.

Now it is your turn to give examples of functions that are not linear. A convincing example has three parts:

a rule that gives exactly one output for each permitted input,
the domain, so it is clear which inputs are permitted,
evidence: three computed points that give two unequal slopes.

Rule 2, $y = x^2$ for all real numbers $x$ , is one example, and you have already seen its evidence. The table below gives three more. Two of them use new notation: $x^3$ means $x \times x \times x$ , and $\frac{12}{x}$ means $12 \div x$ . Division by $0$ is undefined, so the input $0$ is not permitted in that last rule. After the table, a worked example builds one more example from a measurement.

Three more functions that are not linear

Rule and domain	Outputs for three inputs	Points	Two slopes
$y = x^2 + 1$ for all real numbers $x$	$0 \times 0 + 1 = 1$ , $1 \times 1 + 1 = 2$ , $2 \times 2 + 1 = 5$	$(0, 1)$ , $(1, 2)$ , $(2, 5)$	$\frac{2 - 1}{1 - 0} = 1$ and $\frac{5 - 2}{2 - 1} = 3$
$y = x^3$ for all real numbers $x$	$0 \times 0 \times 0 = 0$ , $1 \times 1 \times 1 = 1$ , $2 \times 2 \times 2 = 8$	$(0, 0)$ , $(1, 1)$ , $(2, 8)$	$\frac{1 - 0}{1 - 0} = 1$ and $\frac{8 - 1}{2 - 1} = 7$
$y = \frac{12}{x}$ for all real numbers $x$ except $0$	$12 \div 1 = 12$ , $12 \div 2 = 6$ , $12 \div 3 = 4$	$(1, 12)$ , $(2, 6)$ , $(3, 4)$	$\frac{6 - 12}{2 - 1} = -6$ and $\frac{4 - 6}{3 - 2} = -2$

Worked example

Give an example from a square

Give an example of a function that is not linear, using a square whose side length is $s$ centimeters. Show evidence that your function is not linear.

Choose a quantity that depends on the side length: the area. A square's area is its side length multiplied by itself. Let A stand for the area in square centimeters.
$A = s \times s = s^2$
State the domain. A side length must be positive, so the permitted inputs are all real numbers greater than 0. Each side length gives exactly one area, so this rule is a function.
$s > 0$
Compute three points by multiplying each chosen side length by itself. Side lengths of 2 cm, 5 cm and 10 cm give the points (2, 4), (5, 25) and (10, 100).
$2 \times 2 = 4, \quad 5 \times 5 = 25, \quad 10 \times 10 = 100$
Find the slope between (2, 4) and (5, 25). The area changes by 21 square centimeters while the side changes by 3 centimeters.
$\frac{25 - 4}{5 - 2} = \frac{21}{3} = 7$
Find the slope between (5, 25) and (10, 100). The area changes by 75 square centimeters while the side changes by 5 centimeters.
$\frac{100 - 25}{10 - 5} = \frac{75}{5} = 15$
Compare the slopes. A linear function gives the same slope between every pair of its points, but 7 and 15 are unequal. No single line passes through all three points.
$7 \neq 15$

Answer

One function that is not linear is the area of a square, $A = s^2$ for side lengths $s > 0$ . Its points $(2, 4)$ , $(5, 25)$ and $(10, 100)$ give the unequal slopes $7$ and $15$ .

Finally, compare how the two kinds of function change. Use equal steps in $x$ , so that the changes in $y$ can be compared fairly.

A linear function $y = mx + b$ changes by the same amount, $m$ , at every step of $1$ . Its outputs rise steadily when $m$ is positive, fall steadily when $m$ is negative, and stay the same when $m = 0$ .
A nonlinear function has no single rate of change: at least two of its slopes differ. For example, steps of $1$ in $x$ gave the changes $1$ , $3$ , $5$ for $y = x^2$ , which grow, and $-6$ , then $-2$ for $y = \frac{12}{x}$ , which shrink in size.

So a nonlinear function can change more slowly than a linear one over some steps and faster over others. The first figure showed this: the samples of $y = x^2$ have smaller outputs than Line A at $x = 0$ , $1$ and $2$ (compare $0$ , $1$ , $4$ with $1$ , $3$ , $5$ ), but a larger output at $x = 3$ ( $9$ compared with $7$ ). The next example looks at this more closely.

Worked example

Compare two patterns of change

Both $y = 4x$ and $y = x^2$ have the domain of all real numbers $x$ . Compare how their outputs change over the inputs $x = 0, 1, 2, 3, 4, 5$ .

The first rule is linear, with m = 4 and b = 0. Compute its outputs by multiplying each input by 4.
$4(0) = 0,\ 4(1) = 4,\ 4(2) = 8,\ 4(3) = 12,\ 4(4) = 16,\ 4(5) = 20$
The second rule is the squaring rule. Compute its outputs by multiplying each input by itself.
$0 \times 0 = 0,\ 1 \times 1 = 1,\ 2 \times 2 = 4,\ 3 \times 3 = 9,\ 4 \times 4 = 16,\ 5 \times 5 = 25$
The inputs rise by equal steps of 1, so the changes can be compared fairly. Subtract neighboring outputs of the linear rule: every change is 4.
$4 - 0 = 4,\ 8 - 4 = 4,\ 12 - 8 = 4,\ 16 - 12 = 4,\ 20 - 16 = 4$
Subtract neighboring outputs of the squaring rule: the changes are all different, and they grow.
$1 - 0 = 1,\ 4 - 1 = 3,\ 9 - 4 = 5,\ 16 - 9 = 7,\ 25 - 16 = 9$
Compare the changes step by step, writing the squaring rule's change first. It is less than 4 in the first two steps and more than 4 in the last three.
$1 < 4,\ 3 < 4,\ 5 > 4,\ 7 > 4,\ 9 > 4$
Now compare the outputs at each input, again writing the squaring rule first. It falls behind while its changes are smaller, then catches up at the input 4 and is ahead at the input 5.
$0 = 0,\ 1 < 4,\ 4 < 8,\ 9 < 12,\ 16 = 16,\ 25 > 20$

Answer

Over these inputs, $y = 4x$ rises by the constant amount $4$ at every step. $y = x^2$ rises by $1$ , $3$ , $5$ , $7$ and $9$ : less than $4$ in the first two steps and more than $4$ in the last three. That is why it equals $y = 4x$ at $x = 4$ and is greater at $x = 5$ .

Takeaway. A linear function has one constant rate of change: its rule can be written as $y = mx + b$ , and every pair of its points gives the slope $m$ . A nonlinear function does not: at least two of its slopes differ. Examples include $y = x^2$ , $y = x^3$ , $y = \frac{12}{x}$ for $x \neq 0$ , and the area of a square. To decide, use exact coordinates rather than the look of a graph. Compare slopes, or compare differences when the $x$ -steps are equal. A table speaks only for the inputs it shows unless you also have a rule.

Next, you will evaluate nonlinear functions: finding their outputs for given inputs.