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7. Quadratic Functions and Models
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Algebra I

7.05 Comparing functions

Lesson
Practice

Introduction

We were introduced to key features of linear functions in lesson  2.04 Characteristics of functions  . We learned about characteristics specific to quadratic functions in  7.01 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^2
133
2612
3927
515125
  1. Compare the two 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
  • vertex
  • axis of symmetry

One way to compare functions is to look at growth rates as the x-values increase over regular intervals.

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.

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.

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
  • vertex
  • axis of symmetry

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