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1A. Polynomial and Rational Functions
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AP Precalculus

1.3 Rates of change in linear and quadratic functions

Lesson
Worksheet
Practice
Lesson

Introduction

Learning objectives

  • 1.3.A Determine the average rates of change for sequences and functions, including linear, quadratic, and other function types.
  • 1.3.B Determine the change in the average rates of change for linear, quadratic, and other function types.

Rates of change in linear functions

A linear function can be represented graphically as a straight line. One of the defining characteristics of a linear function is its constant rate of change. This is beacause the rate of change is always the same over every interval of the function. We call this the slope.

Linear function

A function that has a constant rate of change.

Slope

The ratio of the change in the vertical direction (y-direction) to change in the horizontal direction (x-direction).

We can see this constant rate of change by looking at the graph of a linear function.

-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y

Looking at the same function represented in a table:

x-3-2-1012
f(x)-5-3-1135

We can see the constant rate of change by finding the first differences between consecutive f\left(x\right) values:

-3-\left(-5\right)=2, \text{ } -1-\left(-3\right)=2, \text{ } 1-\left(-1\right)=2, \text{ } 3-\left(1\right)=2, \text{ } 5-\left(3\right)=2

These first differences are constant which shows the constant rate of change.

We can use the following formula to calculate the rate of change (or slope) of a linear function:

\displaystyle m= \dfrac{\text{change in }y}{\text{change in }x}=\dfrac{y_2-y_1}{x_2-x_1}
\bm{m}
slope
\bm{\left(x_1,y_1\right)}
a point on the line
\bm{\left(x_2,y_2\right)}
a second point on the line

This formula is very similar to the average rate of change:\dfrac{f(b)-f(a)}{b-a}However, we don't have to take into account the interval a \leq x \leq b since the rate of change of a linear function is constant.

Examples

Example 1

The function f is a linear function with f(6) − f(2) = 20.

a

What is the rate of change of f?

Worked Solution
Create a strategy

We can use the formula: m=\dfrac{f(b)-f(a)}{b-a}

Apply the idea
\displaystyle m\displaystyle =\displaystyle \dfrac{(f(6) - f(2))}{ (6 - 2)}Substitute b=6 and a=2
\displaystyle =\displaystyle \dfrac{20}{6-2}Substitute (f(6) - f(2)) = 20
\displaystyle =\displaystyle \dfrac{20}{4}Evaluate the difference
\displaystyle =\displaystyle 5Simplify
b

Find f(10) - f(6).

Worked Solution
Create a strategy

We can use the formula: m=\dfrac{f(b)-f(a)}{b-a} and use the value of m from part (a).

Apply the idea
\displaystyle 5\displaystyle =\displaystyle \dfrac{(f(10) - f(6))}{ (10 - 6)} Substitute m=5,\,b=10 and a=6
\displaystyle 5\displaystyle =\displaystyle \dfrac{(f(10) - f(6))}{ 4}Simplify the denominator
\displaystyle 20\displaystyle =\displaystyle (f(10) - f(6))Multiply both sides by 4

So, f(10) - f(6) = 20.

c

Find \dfrac{f(27)-f(14)}{5}.

Worked Solution
Apply the idea
\displaystyle 5\displaystyle =\displaystyle \dfrac{f(27)-f(14)}{(27-14)}Substitute m = 5,\,b=27,\, and a=14
\displaystyle 5\displaystyle =\displaystyle \dfrac{(f(27) - f(14))}{13}Simplify the denominator
\displaystyle 65\displaystyle =\displaystyle (f(27) - f(14))Multiply both sides by 13

Now, we want to find \dfrac{(f(27) - f(14))}{5}

\displaystyle \dfrac{(f(27) - f(14))}{5}\displaystyle =\displaystyle 13Divide the result by 5

Example 2

A car rental company charges a daily rate for renting a car. The following table shows the total cost (in dollars) of renting a car for a certain number of days.

\text{Number of Days} (x)\text{Total Cost} (y)
140
3100
5160
7220
9280
a

What is the average rate of change of the total cost between x = 1 and x = 5?

Worked Solution
Create a strategy

We can use the formula m=\dfrac{y_2-y_1}{x_2-x_1}, where x_1 = 1 and x_2 = 5.

Apply the idea
\displaystyle m\displaystyle =\displaystyle \dfrac{160 - 40}{ 5 - 1}Substitute the values for y_{2},\,y_{1},\,x_{2},\, and x_{1}
\displaystyle =\displaystyle \dfrac{120}{4}Evaluate the difference
\displaystyle =\displaystyle \$30 \text{ per day}Simplify
b

Graph the points from the table.

Worked Solution
Create a strategy

Plot the points on the coordinate plane.

Apply the idea
1
2
3
4
5
6
7
8
9
x
20
40
60
80
100
120
140
160
180
200
220
240
260
280
y
c

Is the rate of change of the cost increasing, decreasing, or staying constant?

Worked Solution
Create a strategy

Analyze the given data (table or graph) to observe any patterns or relationships between the number of days and the total cost.

Apply the idea

The rate of change of the cost is staying constant. This is because the relationship between the number of days and the total cost is linear, as observed from the graph of the points. In a linear relationship, the rate of change remains constant throughout.

In this example, we calculated the average rate of change to be \$30 per day, which represents the constant rate at which the total cost increases with respect to the number of days. So, for every additional day, the total cost increases by a constant amount of \$30.

Idea summary

The rate of change of a linear function is its slope which can be calculated using the formula:

\displaystyle m= \dfrac{\text{change in }y}{\text{change in }x}=\dfrac{y_2-y_1}{x_2-x_1}
\bm{m}
slope
\bm{\left(x_1,y_1\right)}
a point on the line
\bm{\left(x_2,y_2\right)}
a second point on the line

Linear functions have a constant first difference.

Rates of change in quadratic functions

Unlike linear functions, which have a constant rate of change, the rates of change in quadratic functions are non-constant, meaning that they vary depending on the input values.

To determine the value by which a quadratic function increases or decreases we look at the first difference, which is the difference between consecutive y-values. Next we can find the difference between consecutive first differences, known as the second difference. If the second difference is a non-zero constant, we have a quadratic relationship.

Consider a table of values for y=x^2.

x-3-2-10123
f\left( x \right)9410149

We can see the first differences are: -5, -3, -1, +1, +3, +5.

Notice that these values are increasing by 2 each time. This means the second differences have a constant value of 2.

We can draw the graph of y=x^2 and see the parabola that is formed. Notice that, for equal x intervals, the y-values are increasing by 2 more for each unit increase in x when x \geq 0.

-4
-3
-2
-1
1
2
3
4
x
-1
1
2
3
4
5
6
7
8
9
10
y

Examples

Example 3

A marble is dropped from the top of a skyscraper, from a height of 1800 feet. The height of the marble, in feet, t seconds after it is dropped is given by the function H(t) = 1800 - 16t^2.

a

Complete the table to show the height of the marble at different times after it is dropped.

t \text{(seconds)}02468
H(t) \text{ feet}
Worked Solution
Create a strategy

Use the given height function H(t) = 1800 - 16t^2, where H(t) represents the height in feet and t represents the time in seconds.

Apply the idea

For each time value, plug in the value of t into the equation and calculate the corresponding height value.

\displaystyle H(0)\displaystyle =\displaystyle 1800-16(0)^2Substitute t=0
\displaystyle =\displaystyle 1800\text{ ft}Evaluate
\displaystyle H(2)\displaystyle =\displaystyle 1800-16(2)^2Substitute t=2
\displaystyle =\displaystyle 1736\text{ ft}Evaluate
\displaystyle H(4)\displaystyle =\displaystyle 1800-16(4)^2Substitute t=4
\displaystyle =\displaystyle 1544\text{ ft}Evaluate
\displaystyle H(6)\displaystyle =\displaystyle 1800-16(6)^2Substitute t=6
\displaystyle =\displaystyle 1224\text{ ft}Evaluate
\displaystyle H(8)\displaystyle =\displaystyle 1800-16(8)^2Substitute t=8
\displaystyle =\displaystyle 776\text{ ft}Evaluate

Complete table:

t \text{(seconds)}02468
H(t) \text{ feet}1800173615441224776
b

Find the exact time t when the marble reaches the ground.

Worked Solution
Create a strategy

We need to solve the equation H(t) = 0 for t, where H(t) = 1800 - 16t^2.

Apply the idea
\displaystyle 0\displaystyle =\displaystyle 1800-16t^2Write the equation
\displaystyle 16t^2\displaystyle =\displaystyle 1800Rearrange the equation
\displaystyle t^2\displaystyle \approx\displaystyle 112.5Divide both sides by 16
\displaystyle t\displaystyle \approx\displaystyle 10.6\text{ seconds}Take the square root of both sides
c

Find the average rate of change in the marble's height during the total length of its drop.

Worked Solution
Create a strategy

Divide the difference in height by the difference in time.

Apply the idea

We need to find the difference in height divided by the difference in time.

\displaystyle \triangle \text{Height}\displaystyle =\displaystyle \text{Final Height} - \text{Initial Height}
\displaystyle =\displaystyle 0-1800Substitute the values
\displaystyle =\displaystyle -1800\text{ ft}Evaluate
\displaystyle \triangle \text{Time}\displaystyle =\displaystyle \text{Final Time} - \text{Initial Time}
\displaystyle =\displaystyle 10.6-0Substitute the values
\displaystyle =\displaystyle 10.6\text{ sec}Evaluate
\displaystyle \text{Average Rate of Change}\displaystyle =\displaystyle \dfrac{\triangle \text{Height}}{\triangle \text{Time}}
\displaystyle =\displaystyle \dfrac{-1800}{10.6}Substitute the values
\displaystyle =\displaystyle -169.81\text{ ft/sec}Evaluate

The average rate of change in the marble's height during the total length of its drop is approximately -169.81 feet per second. The negative sign indicates that the height is decreasing as the marble falls.

d

Complete the table with the average rate of change of the height of the marble.

\text{Time Interval}0\leq t\leq 22\leq t\leq 44\leq t\leq 66\leq t\leq 8
\text{Average rate of change of } \\H \text{ over that interval}
Worked Solution
Create a strategy

We will calculate the height of the marble at the beginning and end of each interval using the function H(t) = 1800 - 16t^2.

The average rate of change will then be the difference between these, divided by the length of the interval.

Apply the idea

Interval 1:

H(0) = 1800 - 16(0)^2 = 1800

H(2) = 1800 - 16(2)^2 = 1736

\displaystyle \dfrac{H(2) - H(0)}{2 - 0}\displaystyle =\displaystyle \dfrac{1736 -1800}{2}Substitute H(0) = 1800 ,\, H(2) = 1736
\displaystyle =\displaystyle -32 \text{ ft}Evaluate

Interval 2:

H(2) = 1800 - 16(2)^2 = 1736

H(4) = 1800 - 16(4)^2 = 1544

\displaystyle \dfrac{H(4) - H(2)}{4 - 2}\displaystyle =\displaystyle \dfrac{1544 - 1736}{2}Substitute H(2) = 1736 ,\, H(4) = 1544
\displaystyle =\displaystyle -96 \text{ ft}Evaluate

Interval 3:

H(4) = 1800 - 16(4)^2 = 1544

H(6) = 1800 - 16(6)^2 = 1224

\displaystyle \dfrac{H(6) - H(4)}{6 - 4}\displaystyle =\displaystyle \dfrac{1224 - 1544}{2}Substitute H(4) = 1544 ,\, H(6) = 1224
\displaystyle =\displaystyle -160 \text{ ft}Evaluate

Interval 4:

H(6) = 1800 - 16(6)^2 = 1224

H(8) = 1800 - 16(8)^2 = 776

\displaystyle \dfrac{H(8) - H(6)}{8 - 6}\displaystyle =\displaystyle \dfrac{776 - 1224}{2}Substitute H(6) = 1224 ,\, H(8) = 776
\displaystyle =\displaystyle -224 \text{ ft}Evaluate

Completed table:

\text{Time Interval}0\leq t\leq 22\leq t\leq 44\leq t\leq 66\leq t\leq 8
\text{Average rate of change of } \\H \text{ over that interval}-32\text{ ft/sec}-96\text{ ft/sec}-160\text{ ft/sec}-224\text{ ft/sec}
e

Plot the points in the table of values from part (a).

Worked Solution
Create a strategy

Plot the points on the coordinate plane from the completed table in part (a).

Apply the idea
1
2
3
4
5
6
7
8
9
x
200
400
600
800
1000
1200
1400
1600
1800
y
f

Is the graph of H concave up or concave down? What does this mean in the context of the problem?

Worked Solution
Apply the idea

The graph of H(t) = 1800 - 16t^2 is concave down. This is because the coefficient of the t^2 term is negative, which results in a downward-opening parabola.

In the context of the problem, a concave-down graph indicates that the rate of change of the height of the marble is not constant. Instead, the marble falls faster and faster as time goes on due to the acceleration of gravity. The concave-down shape of the graph visually represents the increasing speed at which the marble is falling over time.

Example 4

Complete the table for the following quadratic function:

x-3-2-1012345
f\left( x \right)0-3-4-35
Worked Solution
Create a strategy

We first want to determine if the minimum or maximum value of the quadratic function is shown in the table. If it is, we can use the property of symmetry to help complete any missing values.

To find any remaining values, we want to find the first and second differences and use these to complete the table.

Apply the idea

In this case since f(-1) \text{ and }f(1) are equal, the minimum value occurs at the x-value directly between -1 and 1. The minimum value occurs at \left(0, -4 \right).

Since the quadratic is symmetric about this point, we can use this to find other points. We are given the point \left(-2, 0 \right) so using symmetry we know \left (2, 0 \right) must also be on the function. In other words, when we move 2 units to the left of x=0, the output is 0 so when we move 2 units to the right of x=0 the output must also be 0. For the same reason we know when x=-3, the output is 5.

x-3-2-1012345
f\left( x \right)50-3-4-305

To find the last two values we want to find the first and second differences. Comparing the output values we have found so far, we can see the first differences are: -5, -3, -1, +3, +5, \cdots. This means the second difference is +2.

From this information, we can find the remaining values. When x=4, the output will be 5+2=7 more than when x=3, and the value for x=5 will be 7+2=9 more than this value.

x-3-2-1012345
f\left( x \right)50-3-4-3051221
Reflect and check

The minimum or maximum will not necessarily be shown in the table, but it is helpful if it is shown due to the symmetry of quadratic relationships.

We could have also noticed that this table follows a similar pattern to the parent quadratic function, f(x)=x^2, but is translated down 4 units. If we noticed this, we could use the rule to complete the missing values.

Idea summary

The rate of change of a quadratic function changes at a linear rate.

The second difference for a quadratic function is constant.

Outcomes

1.3.A

Determine the average rates of change for sequences and functions, including linear, quadratic, and other function types.

1.3.B

Determine the change in the average rates of change for linear, quadratic, and other function types.

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