Comparing Rates of Change and Initial Values | Grade 8 Math | LevelKite

You already know how to study one function at a time. Now you will put two functions side by side and ask two separate questions: How fast does each one change? and Where does each one start?

First, a quick reminder of the words you need. A function gives exactly one output for each input it allows. Two different inputs may share an output, but one input never has two outputs. The set of allowed inputs is the function's domain.

Every function in this lesson is a linear function: you can write its rule as $y = mx + b$ , where $x$ is the input, $y$ is the output, and $m$ and $b$ are fixed numbers. Its graph lies on a straight line that is not vertical. A vertical line is never the graph of a function: one input would have many outputs, and its slope (change in $y$ divided by change in $x$ ) is undefined because the change in $x$ along it is 0.

The figure below shows two linear functions on one coordinate plane. The horizontal axis shows the input $x$ and the vertical axis shows the output $y$ , so the point $(2, 5)$ means the input 2 gives the output 5. Both functions allow every real number as an input, and the figure shows inputs from 0 to 4.

Line A is the graph of Function A, $y = 2x + 1$ . Its labelled points are $(0, 1)$ and $(2, 5)$ .
Line B is the graph of Function B, $y = x + 6$ . Its labelled points are $(0, 6)$ and $(2, 8)$ .

Follow each line from input 0 to input 2. Function A's output goes from 1 to 5, a change of 4 while the input takes 2 steps of 1. A linear function changes by the same amount on each of those steps, so each step adds $4 \div 2 = 2$ . Function B's output goes from 6 to 8, a change of 2 over the same 2 steps, so each step adds $2 \div 2 = 1$ . This change in output for each increase of 1 in the input is the rate of change.

Now look at input 0. Function A's output there is 1, and Function B's is 6. The output at input 0 is the initial value.

So Function A changes faster, but Function B starts higher.

On Line A,

2(0) + 1 = 1

and

2(2) + 1 = 5

. On Line B,

0 + 6 = 6

and

2 + 6 = 8

. Each labelled point shows an input and its output.

Key concept

Two questions, two separate comparisons

These rules are for linear functions. Take any two input-output pairs $(x_1, y_1)$ and $(x_2, y_2)$ with different inputs. The rate of change is the change in output divided by the change in input. Every such pair gives the same number: the slope of the function's line, and the $m$ in the rule $y = mx + b$ .

The initial value is the output at input 0, so a function has one only when 0 is in its domain. In $y = mx + b$ it is $b$ . On a graph it is the $y$ -coordinate of the point where $x = 0$ .

Compare rate with rate and initial value with initial value, in the same units for both functions. On a graph, use exact labelled points, not how steep or high a line looks. One comparison does not decide the other.

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

Function D is linear. The table shows three of its input-output pairs.

Input x	Output y
$1$	$9$
$3$	$13$
$5$	$17$

Worked example

Comparing rates of change from a rule and a table

Function C has the rule $y = 3x + 2$ . Function D is linear, and the table above shows three of its input-output pairs: $(1, 9)$ , $(3, 13)$ and $(5, 17)$ . Both functions allow every real number as an input. Which function has the greater rate of change?

Find Function C's rate of change from its rule. Each time x goes up by 1, the term 3x goes up by 3 and the added 2 stays the same, so the output goes up by 3. Confirm this with two inputs: input 0 gives 3(0) + 2 = 2, and input 1 gives 3(1) + 2 = 5.
$\dfrac{5 - 2}{1 - 0} = \dfrac{3}{1} = 3$
Find Function D's rate of change from two rows of its table. The rows with inputs 1 and 3 have different inputs, so the change in input is not 0 and you can divide the change in output by it.
$\dfrac{13 - 9}{3 - 1} = \dfrac{4}{2} = 2$
Check with another pair of rows, the ones with inputs 3 and 5. The problem tells you Function D is linear, so every pair of rows must give the same rate, and this rate holds for all of its inputs, not only the three rows shown.
$\dfrac{17 - 13}{5 - 3} = \dfrac{4}{2} = 2$
Compare the two rates. Each one is the change in output for each increase of 1 in the input, so you can compare them directly.
$3 > 2$

Answer

Function C has the greater rate of change. Its output goes up by $3$ for each increase of $1$ in the input, and Function D's output goes up by $2$ .

Function F is linear. The table shows three of its input-output pairs.

Input x	Output y
$2$	$7$
$4$	$13$
$6$	$19$

Line E is the graph of Function E,

y = x + 4

, shown for inputs

0

to

6

. Its labelled points are

(0, 4)

and

(2, 6)

.

Worked example

Comparing initial values from a graph and a table

Function E is linear, and its graph is Line E in the figure above. The horizontal axis shows the input $x$ and the vertical axis shows the output $y$ . Function F is linear, and the table above shows three of its input-output pairs: $(2, 7)$ , $(4, 13)$ and $(6, 19)$ . Both functions allow every real number as an input. Which function has the greater initial value?

Read Function E's initial value from its graph. Line E has two labelled points, and the one with input 0 is (0, 4). It tells you the input 0 gives the output 4, so Function E's initial value is 4. This agrees with the rule in the figure's caption, y = x + 4.
$0 + 4 = 4$
Function F's table has no row with input 0, so find its rate of change first. The rows with inputs 2 and 4 have different inputs, so divide the change in output by the change in input.
$\dfrac{13 - 7}{4 - 2} = \dfrac{6}{2} = 3$
Step back from input 2 to input 0. Function F is linear and allows every real number as an input, so 0 is in its domain and the rate of 3 also holds between inputs 0 and 2. The input goes down by 2, so the output goes down by 2 times 3, which is 6.
$7 - 2 \cdot 3 = 7 - 6 = 1$
Check the result. Function F is linear, so an initial value of 1 and a rate of change of 3 give its rule, y = 3x + 1. Test this rule on every row of the table.
$3(2) + 1 = 7, \quad 3(4) + 1 = 13, \quad 3(6) + 1 = 19$
Compare the two initial values. Each one is an output at input 0, so you can compare them directly.
$4 > 1$

Answer

Function E has the greater initial value. Its output at input $0$ is $4$ , and Function F's output at input $0$ is $1$ .

Line G shows Tank G for

0 \le x \le 8

, where

x

is minutes and

y

is gallons. Its labelled points are

(0, 6)

and

(4, 12)

.

Worked example

Comparing two tanks: fill rate and starting amount

Two tanks fill with water for 8 minutes. Let $x$ be the number of minutes since filling began, with $0 \le x \le 8$ , and let $y$ be the gallons of water in a tank. Tank G's function is linear, and its graph is Line G in the figure above. Tank H follows the rule $y = 0.5x$ . Which tank fills faster, and which tank starts with more water?

Find Tank G's rate of change from Line G. The horizontal axis shows minutes and the vertical axis shows gallons, so (0, 6) means 6 gallons at 0 minutes and (4, 12) means 12 gallons at 4 minutes. The inputs 0 and 4 are different, so divide the change in gallons by the change in minutes. The result is exactly 1.5.
$\dfrac{12 - 6}{4 - 0} = \dfrac{6}{4} = 1.5$
Find Tank G's initial value. The input 0 is in the domain, and the labelled point (0, 6) gives 6 gallons at 0 minutes. Check both results: Tank G's function is linear, so a rate of 1.5 and an initial value of 6 give the rule y = 1.5x + 6, and this rule must fit the other labelled point, (4, 12).
$1.5(4) + 6 = 6 + 6 = 12$
Read both features of Tank H from its rule, y = 0.5x. Each extra minute adds 0.5 gallons, so its rate of change is 0.5. At 0 minutes the tank holds 0.5(0) = 0 gallons, so its initial value is 0.
$0.5(0) = 0$
Compare rate with rate and initial value with initial value. Both rates are in gallons per minute and both initial values are in gallons, so you can compare each pair directly.
$1.5 > 0.5 \quad \text{and} \quad 6 > 0$

Answer

Tank G fills faster: $1.5$ gallons per minute compared with Tank H's $0.5$ gallons per minute. Tank G also starts with more water: $6$ gallons compared with Tank H's $0$ gallons.

To compare two linear functions, ask two separate questions and answer each with exact values.

How fast does each one change? Find each rate of change: the change in output divided by the change in input, using two pairs with different inputs.
Where does each one start? Find each initial value: the output at input 0, when 0 is in the domain. A table row gives the start only when its input is 0.

You can find both from a rule, from a table of a function you know is linear, or from exact labelled points on a graph. Sometimes each function has the greater value in one comparison, as Functions A and B did. Sometimes one function has the greater value in both, as Tank G did. That is why you always answer the two questions separately.

Next, you will compare properties of two functions represented in different ways.