Comparing Functions Across Equations, Tables and Graphs | Grade 8 Math | LevelKite

A function is a rule that gives each permitted input exactly one output. Different inputs may share an output, but one input never has two outputs. The set of permitted inputs is the function's domain.

You can show one function in different ways: as an equation, as a table of input-output pairs, or as a graph. These are three views of the same rule, so they always agree. Take $y = 2x + 1$ , where $x$ is the input, $y$ is the output, and $x$ can be any real number. The input $3$ gives the output $2 \cdot 3 + 1 = 6 + 1 = 7$ , so $(3, 7)$ is one of its input-output pairs. A table of this function that includes the input $3$ lists $7$ beside it, and its graph passes through the point $(3, 7)$ .

In this lesson, you will compare two functions that are shown in different ways. You can't set an equation next to a picture and compare them directly. Instead, you pull the same piece of information out of each one, and then you compare.

Every function in this lesson is a linear function. Its rule can be written as $y = mx + b$ , where $m$ and $b$ are fixed numbers, and its graph is a straight line that is not vertical. Two properties describe it.

The rate of change is how much the output changes each time the input goes up by $1$ . For a linear function, it is the same everywhere.
The initial value is the output when the input is $0$ , as long as $0$ is in the domain. You have already compared initial values in an earlier lesson.

Look at $y = 2x + 1$ again. When $x$ goes up by $1$ , the term $2x$ goes up by $2$ and the added $1$ stays the same, so the output goes up by $2$ . The rate of change is $2$ , the number that multiplies $x$ . When $x = 0$ , the output is $2 \cdot 0 + 1 = 0 + 1 = 1$ . The initial value is $1$ , the number that is added on its own.

A table and a graph hold these same two numbers. You just find them in a different way.

Key concept

Same property, different forms

To compare two linear functions shown in different ways, find the same property in each one. Then compare the two numbers.

Rate of change. In an equation $y = mx + b$ , it is $m$ . In a table or on a graph, take two input-output pairs $(x_1, y_1)$ and $(x_2, y_2)$ with different inputs. Divide the change in output by the change in input, as the formula shows.
Initial value, when $0$ is in the domain. In $y = mx + b$ , it is $b$ . In a table, it is the output beside the input $0$ ; if that row is missing, work back to $0$ using the rate. On a graph, it is the $y$ -coordinate where the line crosses the $y$ -axis.

A table shows only the pairs it lists. Use its rate for other inputs only when you know the function is linear.

\text{rate of change} = \frac{y_2 - y_1}{x_2 - x_1}, \quad x_2 \neq x_1

Function B: a linear function whose domain is all real numbers

Input x	Output y
$0$	$5$
$1$	$7$
$2$	$9$
$3$	$11$

Worked example

Equation and table

Function A is $y = 4x + 3$ , where $x$ can be any real number. Function B is a linear function whose domain is also all real numbers. The table above lists four of its input-output pairs. Which function has the greater rate of change? Which has the greater initial value?

Read Function A's rate of change from its equation. Each time x goes up by 1, the term 4x goes up by 4 and the added 3 stays the same, so the output goes up by 4. The rate is the number that multiplies x.
$\text{rate of A} = 4$
Find Function B's rate of change from two table rows with different inputs, (0, 5) and (1, 7). Divide the change in output by the change in input to get the change for each 1 unit of input. B is linear, so every pair of rows gives this same rate. The rows (2, 9) and (3, 11) confirm it.
$\frac{7 - 5}{1 - 0} = \frac{2}{1} = 2 \qquad \frac{11 - 9}{3 - 2} = \frac{2}{1} = 2$
Find each initial value, the output when the input is 0. The input 0 is in both domains. For A, substitute x = 0 into the equation. For B, read the output in the table row where x is 0.
$\text{A: } 4 \cdot 0 + 3 = 0 + 3 = 3 \qquad \text{B: } 5$
Compare the same property from each function. For the rates of change, 4 is greater than 2. For the initial values, 5 is greater than 3.
$4 > 2 \qquad 5 > 3$

Answer

Function A has the greater rate of change, $4$ compared with $2$ : its output rises by more each time $x$ goes up by $1$ . Function B has the greater initial value, $5$ compared with $3$ : it has the greater output at $x = 0$ .

A graph holds the same two properties. The rate of change of a linear function is the slope of its line: the change in $y$ divided by the change in $x$ between two different points on the line. The graph of a function is never a vertical line. A vertical line would give one input many outputs. Its slope is also undefined, because the change in $x$ would be $0$ and you cannot divide by $0$ .

To find a slope, use points whose coordinates you know exactly. Don't judge by how steep the line looks, because the look depends on the scale of the figure and the coordinates do not.

In the figure below, the horizontal axis shows the input $x$ and the vertical axis shows the output $y$ . Each grid square is $1$ unit wide and $1$ unit tall. Line C is the graph of Function C, a linear function whose domain is all real numbers. Line C passes through the marked points $(0, 1)$ and $(2, 7)$ . The point $(0, 1)$ lies on the $y$ -axis, so it shows the output when the input is $0$ .

Line C is the graph of Function C. It passes through

(0, 1)

and

(2, 7)

, so

y

goes up by

6

while

x

goes up by

2

.

Function D: a linear function whose domain is all real numbers

Input x	Output y
$2$	$3$
$4$	$7$
$6$	$11$

Worked example

Graph and table

Function C is the linear function graphed as Line C in the figure above, and its domain is all real numbers. Function D is a linear function whose domain is also all real numbers. The table just above lists three of its input-output pairs. Which function has the greater rate of change? Which has the greater initial value?

Find Function C's rate of change from the two marked points on Line C, (0, 1) and (2, 7). Line C is not vertical, so the two inputs differ and you can divide the change in y by the change in x.
$\frac{7 - 1}{2 - 0} = \frac{6}{2} = 3$
Find Function D's rate of change from the rows (2, 3) and (4, 7). Divide the change in output by the change in input. D is linear, so the rows (4, 7) and (6, 11) must give the same rate, and they do.
$\frac{7 - 3}{4 - 2} = \frac{4}{2} = 2 \qquad \frac{11 - 7}{6 - 4} = \frac{4}{2} = 2$
Read Function C's initial value from the graph. Line C crosses the y-axis at (0, 1), so the output is 1 when the input is 0.
$\text{initial value of C} = 1$
Function D's table has no row for x = 0. The first row is (2, 3), so 3 is the output at input 2, not the initial value. D is linear with rate 2, and 0 is in its domain. Going back from input 2 to input 0 is 2 units, so the output drops by 2 for each unit: 2 times 2, which is 4.
$3 - 2 \cdot 2 = 3 - 4 = -1$
Compare the same property from each function. For the rates, 3 is greater than 2. For the initial values, a positive number is greater than a negative number, so 1 is greater than -1.
$3 > 2 \qquad 1 > -1$

Answer

Function C has the greater rate of change, $3$ compared with $2$ . Function C also has the greater initial value, $1$ compared with $-1$ .

Sometimes you need to compare outputs at one chosen input, and a table or graph does not show that input. Don't estimate from the picture. When you know a function is linear, with both $0$ and your chosen input in its domain, you can build its rule from its rate of change and its initial value. Then you substitute.

In the figure below, the horizontal axis shows the input $x$ and the vertical axis shows the output $y$ . Each grid square is $1$ unit wide and $1$ unit tall. Line F is the graph of Function F, a linear function whose domain is all real numbers. Line F passes through the marked points $(0, 0)$ and $(2, 8)$ . The line leaves the top of the figure before it reaches $x = 5$ , so the picture cannot show you the output there.

Line F is the graph of Function F. It passes through

(0, 0)

and

(2, 8)

, so

y

goes up by

8

while

x

goes up by

2

.

Worked example

Equation and graph

Function E is $y = 3x + 4$ , where $x$ can be any real number. Function F is the linear function graphed as Line F in the figure above, and its domain is all real numbers. Which function has the greater output at $x = 5$ ? Is either function proportional, meaning its rule has the form $y = kx$ for a fixed nonzero number $k$ ?

Find Function F's rate of change from the marked points (0, 0) and (2, 8) on Line F. The line is not vertical, so the inputs differ and you can divide.
$\frac{8 - 0}{2 - 0} = \frac{8}{2} = 4$
Read F's initial value. Line F crosses the y-axis at (0, 0), so the output is 0 when the input is 0. F is linear, so its rule is its rate times x, plus its initial value.
$y = 4x + 0 = 4x$
Substitute x = 5 into each rule. You can do this because 5 is in both domains, even though the figure does not show Line F at that input.
$\text{E: } 3 \cdot 5 + 4 = 15 + 4 = 19 \qquad \text{F: } 4 \cdot 5 = 20$
Compare the two outputs at the same input, x = 5.
$20 > 19$
Test each function. A rule of the form y = kx gives a line through the origin (0, 0), and y divided by x equals k for every x that is not 0. F's rule is y = 4x, so F is proportional with k = 4. E's initial value is 4, not 0, so its graph misses the origin. E is linear but not proportional.
$y = 4x \quad \text{has the form} \quad y = kx \quad \text{with } k = 4$

Answer

At $x = 5$ , Function F has the greater output, $20$ compared with $19$ . Function F is proportional. Function E is linear but not proportional.

Takeaway: an equation, a table and a graph are different views of a function, so you can compare functions across them. Find the same property in each one, such as the rate of change, the initial value or the output at a chosen input. Use exact values, and use what you know about each function's domain and whether it is linear. Then compare the numbers. Next, you will use these comparisons in word problems and justify which function grows faster.