Topics
10. Quadratic Functions and Models
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United States of America
Grades 9-12

10.05 Comparing functions

Lesson
Practice

Introduction

We were introduced to key features of linear functions in lesson  3.04 Characteristics of functions  . We continued learning about characteristics of functions and began comparing functions in lesson  5.01 Exponential functions  and lesson  6.06 Comparing linear and exponential functions  . We learned about characteristics specific to quadratic functions in  10.01 Characteristics of quadratic functions  and will use key features to compare various functions represented in different ways in this lesson.

Comparing functions

Exploration

Consider the table below:

xy=3xy=3x^2y=3^x
1333
26129
392727
515125243
  1. Compare the three functions and how they change as x increases.

The way a function is represented can affect the characteristics we are able to identify for the function. Different representations can highlight or hide certain characteristics. Remember that key features of functions include:

  • domain and range
  • x- and y-intercepts
  • maximum or minimum value(s)
  • average rate of change over various intervals
  • end behavior
  • positive and negative intervals
  • increasing and decreasing intervals
  • asymptote(s)
  • vertex
  • axis of symmetry

One way to compare functions is to look at growth rates as the x-values increase over regular intervals. In order to compare the growth rates of quadratics with those of exponential or linear functions, we will examine only the increasing interval of a quadratic function.

When the leading coefficient of the quadratic equation is positive, the parabola opens upward. In this case, we know y increases at an increasing rate as x approaches infinity.

Since a linear function increases at a constant rate and the quadratic function increases at an increasing rate as x increases, eventually the quadratic function will increase faster than the linear function.

Next, we need to examine how an exponential growth function compares to the increasing portion of the quadratic function, since both functions increase at an increasing rate. Consider a situation where we compare the increasing interval of the quadratic function g(x) with a positive leading coefficient, to an exponential growth function h(x), as shown in the graph.

-2
-1
1
2
3
4
5
6
x
10
20
30
y

Notice starting at x=0, g(x) is greater than h(x) and is increasing at a greater rate. But, as x continues to increase, the quadratic function g(x) is increasing at a slower rate than the exponential function, and eventually the exponential function will overtake the quadratic function.

An exponential growth function will always exceed a linear or quadratic growth function as values of x become larger.

Examples

Example 1

Consider the functions shown below. Assume that the domain of f is all real numbers.

  • Function 1:

    x-1012345
    f\left(x\right)-3.75-2-0.251.53.2556.75

  • Function 2:

    -8
    -6
    -4
    -2
    2
    4
    6
    8
    x
    -4
    -2
    2
    4
    6
    8
    10
    12
    y
a

Determine which function has a higher y-intercept.

Worked Solution
Create a strategy

Remember that the y-intercept of a function occurs when x=0. We can use this to evaluate the y-intercept of f and identify the y-intercept of g.

Apply the idea

For f, we can see from the table that f\left(0\right) = -2.

For g, we can see from the graph that g\left(0\right) = -3.

So the y-intercept of f is the point \left(0, -2\right) and the y-intercept of g is the point \left(0, -3\right), and therefore f has a higher y-intercept.

b

Find the average rate of change for each function over the following intervals:

  • 0 \leq x \leq 1
  • 1 \leq x \leq 4
  • 4 \leq x \leq 5
Worked Solution
Create a strategy

For Function 1, we can find the values of f\left(0\right), f\left(1\right), f\left(4\right), and f\left(5\right) from the table of values. For Function 2 we will need to look at the graph and estimate the values of g\left(0\right), g\left(1\right), g\left(4\right), and g\left(5\right).

Apply the idea

First, we can consider the interval 0 \leq x \leq 1.

For Function 1, using the table of values, we can see that f\left(0\right)=-2 and f\left(1\right)=-0.25.

So, the average rate of change of Function 1 over 0 \leq x \leq 1 is:\dfrac{f(1)-f(0)}{1-0}=\dfrac{-0.25-\left(-2\right)}{1-0}=\dfrac{1.75}{1}=1.75

For Function 2, using the graph, we can see that g\left(0\right)=-3 and g\left(1\right)=-4.

So, the average rate of change of Function 2 over 0 \leq x \leq 1 is:\dfrac{g(1)-g(0)}{1-0}=\dfrac{-4-\left(-3\right)}{1-0}=-\dfrac{1}{1}=-1

Next, we can consider the interval 1 \leq x \leq 4.

For Function 1, using the table of values, we can see that f\left(1\right)=-0.25 and f\left(4\right)=5.

So, the average rate of change of Function 1 over 1 \leq x \leq 4 is:\dfrac{f(4)-f(1)}{4-1}=\dfrac{5-\left(-0.25\right)}{4-1}=\dfrac{5.25}{3}=1.75

For Function 2, using the graph, we can see that g\left(1\right)=-4 and g\left(4\right)=5.

So, the average rate of change of Function 2 over 1 \leq x \leq 4 is:\dfrac{g(4)-g(1)}{4-1}=\dfrac{5-\left(-4\right)}{4-1}=\dfrac{9}{3}=3

Lastly, we can consider the interval 4 \leq x \leq 5.

For Function 1, using the table of values, we can see that f\left(4\right)=5 and f\left(5\right)=6.75.

So, the average rate of change of Function 1 over 4 \leq x \leq 5 is:\dfrac{f(5)-f(4)}{5-4}=\dfrac{6.75-5}{5-4}=\dfrac{1.75}{1}=1.75

For Function 2, using the graph, we can see that g\left(4\right)=5 and g\left(5\right)=12.

So, the average rate of change of Function 2 over 4 \leq x \leq 5 is:\dfrac{g(5)-g(4)}{5-4}=\dfrac{12-5}{5-4}=\dfrac{7}{1}=7

Reflect and check

We can notice from the table of values that f\left(x\right) is increasing at a constant rate. This means that regardless of the interval we consider, the average rate of change remains the same.

c

Using part (b), determine which function will be greater as x approaches positive infinity.

Worked Solution
Create a strategy

We can consider the average rate of change calculated in part (b) to determine which function will be greater for large values of x.

Apply the idea

From part (b), we saw that f\left(x\right) was a linear function with a constant rate of change of 1.75. We calculated that g\left(x\right) had an increasing rate of change as x increased. So, we can see that Function 2 will be far greater than Function 1 as x approaches positive infinity.

Reflect and check

A concave up quadratic function will grow faster than a linear function as x approaches positive infinity.

Example 2

Consider the functions below:

  • Function 1: f(x) = \begin{cases} -x^{2}+5, & x \leq 0 \\ -\dfrac{1}{2}x-3, & 0<x \end{cases}
  • Function 2:
    -4
    -3
    -2
    -1
    1
    2
    3
    4
    x
    5
    10
    15
    20
    g(x)
a

Determine which function is increasing if x>0.

Worked Solution
Create a strategy

Graph Function 1 to visualize where the function is increasing and use the graph of Function 2 to visualize where the function is increasing.

-8
-6
-4
-2
2
4
6
8
x
-8
-6
-4
-2
2
4
6
8
y
Apply the idea

If x>0, Function 2 is increasing.

Reflect and check

We also could have just looked at the equation for Function 1 when x>0, and identified it as a linear function with a slope of -\dfrac{1}{2}, meaning the function would be decreasing over that interval.

b

Determine which function has a greater magnitude in its rate of change over the interval 1 \leq x \leq 2.

Worked Solution
Create a strategy

Calculate the average rate of change for each function over the interval 1 \leq x \leq 2. Since Function 1 will be -\dfrac{1}{2}x - 3 over 1 \leq x \leq 2, calculate the average rate of change using that expression. For Function 2, we will need to look at the graph and estimate the values of f(1) and f(2).

Apply the idea

For Function 1, we can evaluate f(2) and f(1):f(2)=-\dfrac{1}{2}(2)-3=-1-3=-4 \\ f(1)=-\dfrac{1}{2}(1)-3=-0.5-3=-3.5

The average rate of change of Function 1 over 1 \leq x \leq 2 is\dfrac{f(2)-f(1)}{2-1}=\dfrac{-4-(-3.5)}{2-1}=-\dfrac{0.5}{1}=-0.5

The average rate of change of Function 2 over 1 \leq x \leq 2 is \dfrac{f(2)-f(1)}{2-1}=\dfrac{20-9}{2-1}=\dfrac{11}{1}=11.

The rate of change for Function 1 is negative, which means the function is decreasing. The rate of change for Function 2 is positive, which means the function is increasing. However, since we want to compare the magnitudes -or steepness- of their rates of change, we need to compare the amount of change without regard to negative or positve values. The ratio of increase, 11, for Function 2 is greater than the ratio of decrease, 0.5, for Function 1. Therefore, Function 2 has a greater magnitude in its rate of change over 1 \leq x \leq 2.

c

Determine whether or not each function has a maximum or minimum value.

Worked Solution
Create a strategy

Function 1 is a piecewise function with a quadratic function and a linear function. Function 2 is an exponential function.

Apply the idea

Function 1 has a maximum point at the point (0,5), which is the vertex of the quadratic portion of the piecewise function.

Function 2 is exponential, and has a horizontal asymptote at y=5. Because the function never actually reaches 5 as x approaches negative infinity, we can not call this a minimum value. There are an infinite number of smaller and smaller fractions that get close to, but never quite reach, 5. As x approaches positive, infinity, the exponential function will also approach positive infinity. Therefore, Function 2 does not have a minimum or maximum value.

Example 3

Consider functions representing three options to earn money one of the following ways:

A figure showing 3 options to earn money. Option 1 shows the statement: You are given 2 dollars each day. Option 2 shows a table with 2 columns titled Days and Total Amount and with 6 rows. The data is as follows: First column: 1, 2, 3, 4, 5,6; Second column: 1 dollar, 4 dollars, 9 dollars, 16 dollars, 25 dollars, 36 dollars. Option 3 shows a first quadrant coordinate plane with the x axis labeled Days and the y axis labeled Total Amount in dollars. The points (1, 2), (2, 4), (3, 8), (4, 16), and (5, 32) are plotted on the graph.

Note: Option 3 starts with \$2 on day one and doubles each day after this.

a

Compare the average rate of change of each function over the intervals 2 \leq x \leq 3 and 4 \leq x \leq 5.

Worked Solution
Create a strategy

To find the average rate of change from a given function over the interval a \leq x \leq b, we can find the change in the value of the dependent variable f(b)-f(a) per change in value in the independent variable b-a, or:

\dfrac{f(b)-f(a)}{b-a}

for each of the options.

Apply the idea

For Option 1, f(2)=2 \cdot 2 = 4 and f(3)=2 \cdot 3=6.

The average rate of change for Option 1 over the interval 2 \leq x \leq 3 is \dfrac{f(3)-f(2)}{3-2} = \dfrac{6-4}{4-3} = \dfrac{2}{1} = 2

For Option 2, f(2)=4 and f(3)=9.

The average rate of change for Option 2 over the interval 2 \leq x \leq 3 is \dfrac{f(3)-f(2)}{3-2} = \dfrac{9-4}{4-3} = \dfrac{5}{1} = 5

For Option 3, f(2)=4 and f(3)=8.

The average rate of change for Option 3 over the interval 2 \leq x \leq 3 is \dfrac{f(3)-f(2)}{3-2} = \dfrac{8-4}{4-3} = \dfrac{4}{1} = 4

Option 2 has the greatest average rate of change over 2 \leq x \leq 3, at \$5 per day.

For Option 1, f(4)=2 \cdot 4 = 8 and f(5)=2 \cdot 5=10.

The average rate of change for Option 1 over the interval 4 \leq x \leq 5 is \dfrac{f(5)-f(4)}{5-4} = \dfrac{10-8}{5-4} = \dfrac{2}{1} = 2

For Option 2, f(4)=16 and f(5)=25.

The average rate of change for Option 2 over the interval 4 \leq x \leq 5 is \dfrac{f(5)-f(4)}{5-4} = \dfrac{25-16}{5-4} = \dfrac{9}{1} = 9

For Option 3, f(4)=16 and f(5)=32.

The average rate of change for Option 3 over the interval 4 \leq x \leq 5 is \dfrac{f(5)-f(4)}{5-4} = \dfrac{32-16}{5-4} = \dfrac{16}{1} = 16

The average rate of change of Option 1 remained \$2 per day over both intervals, while the average rates of change of Options 2 and 3 both increased from the first to the second intervals. Option 3 has the greatest rate of change over 4 \leq x \leq 5, at \$16 per day.

b

Find the equation that represents each option, where x is the number of days that have passed.

Worked Solution
Create a strategy

For each option, we can consider how the total amount of money changes as the days progress and derive an equation to represent the relationship.

Apply the idea

For Option 1, we saw in part (a) that we had a constant rate of change regardless of the interval we considered. So, Option 1 can be represented by the linear function, f\left(x\right)=2x.

Now, observing the table of values for Option 2, we can see that the total amount is just the square of the number of days passed. So, Option 2 can be represented by the function f\left(x\right)=x^2.

Finally, the relationship for Option 3 is represented in the graph, but also described to us. Since we are told that the function starts at \$2 and is doubled each day, we can see that Option 3 is just represented by the function f\left(x\right)=2^x.

Reflect and check

If the relationship between the days passed and the total amount weren't directly obvious in Option 2, we could have tested the data provided in the table to rule out a linear or exponential relationship.

For a linear relationship, the average rate of change between any two points must be equal. We already discovered in part (a) that this wasn't true for Option 2. So, we could have then tested if it represented an exponential relationship.

For an exponential relationship, the ratio of between two points, a unit apart, must be equal. We can see that for Option 2, \dfrac{4}{1} \neq \dfrac{9}{4}.

Therefore, we could see that Option 2 represented neither a linear or exponential relationship.

c

Find the value of each option at 8 days, 12 days, and 14 days.

Worked Solution
Create a strategy

Construct a table of values with the amounts of money gained with each option.

Apply the idea
DaysOption 1 TotalOption 2 TotalOption 3 Total
1\$2\$1\$2
2\$4\$4\$4
3\$6\$9\$8
4\$8\$16\$16
5\$10\$25\$32
6\$12\$36\$64
7\$14\$49\$128
8\$16\$64\$256
9\$18\$81\$512
10\$20\$100\$1\,024
11\$22\$121\$2\,048
12\$24\$144\$4\, 096
13\$26\$169\$8\,192
14\$28\$196\$16 \,384

At 8 days, Option 1 will make \$16, Option 2 will make \$64, and Option 3 will make \$256.

At 12 days, Option 1 will make \$24, Option 2 will make \$144, and Option 3 will make \$4\,096.

At 14 days, Option 1 will make \$28, Option 2 will make \$196, and Option 3 will make \$16\,384.

Reflect and check

We could calculate the total amount of money on days 8, 12 and 14 using the functions found in part (b), instead of constructing a table.

d

Determine which option will be greater for larger and larger values of x.

Worked Solution
Create a strategy

Use the table comparison from part (b) to determine which option will be greater for larger and larger values of x.

Apply the idea

As x gets larger and larger, we can see that Option 3, the exponential option, will be far greater than Options 1 or 2.

Reflect and check

An exponential function will always exceed a linear or quadratic function as values of x become larger.

Idea summary

It is important to be able to compare the key features of functions whether they are represented in similar or different ways:

  • domain and range
  • x- and y-intercepts
  • maximum or minimum value(s)
  • average rate of change
  • end behavior
  • positive and negative intervals
  • increasing and decreasing intervals
  • asymptote(s)
  • vertex
  • axis of symmetry

Outcomes

F.IF.C.9

Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions).

F.BF.A.1

Write a function that describes a relationship between two quantities.

F.BF.A.1.A

Determine an explicit expression, a recursive process, or steps for calculation from a context.

F.LE.A.3

Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function.

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