Learning Objectives

In this section, you will:

5.7.1 – Finding the Domain and Range of a Quadratic Function

5.7.2 – Determining the Maximum and Minimum Values of Quadratic Functions

5.7.1 – Finding the Domain and Range of a Quadratic Function

Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y-values greater than or equal to the y-coordinate at the turning point or less than or equal to the y-coordinate at the turning point, depending on whether the parabola opens up or down.

Domain and Range of a Quadratic Function

The domain of any quadratic function is all real numbers unless the context of the function presents some restrictions.

The range of a quadratic function written in general form [latex]\,f\left(x\right)=a{x}^{2}+bx+c\,[/latex] with a positive [latex]\,a\,[/latex] value is [latex]\,f\left(x\right)\ge f\left(-\frac{b}{2a}\right),\,[/latex] or [latex]\,\left[f\left(-\frac{b}{2a}\right),\infty \right);\,[/latex] the range of a quadratic function written in general form with a negative [latex]\,a\,[/latex] value is [latex]\,f\left(x\right)\le f\left(-\frac{b}{2a}\right),\,[/latex] or [latex]\,\left(-\infty ,f\left(-\frac{b}{2a}\right)\right].[/latex]

The range of a quadratic function written in standard form [latex]\,f\left(x\right)=a{\left(x-h\right)}^{2}+k\,[/latex] with a positive [latex]\,a\,[/latex] value is [latex]\,f\left(x\right)\ge k;\,[/latex] the range of a quadratic function written in standard form with a negative [latex]\,a\,[/latex] value is [latex]\,f\left(x\right)\le k.[/latex]

How To

Given a quadratic function, find the domain and range.

  1. Identify the domain of any quadratic function as all real numbers.
  2. Determine whether [latex]\,a\,[/latex] is positive or negative. If [latex]\,a\,[/latex] is positive, the parabola has a minimum. If [latex]\,a\,[/latex] is negative, the parabola has a maximum.
  3. Determine the maximum or minimum value of the parabola, [latex]\,k.[/latex]
  4. If the parabola has a minimum, the range is given by [latex]\,f\left(x\right)\ge k,\,[/latex] or [latex]\,\left[k,\infty \right).\,[/latex] If the parabola has a maximum, the range is given by [latex]\,f\left(x\right)\le k,\,[/latex] or [latex]\,\left(-\infty ,k\right].[/latex]

Example 1 – Finding the Domain and Range of a Quadratic Function

Find the domain and range of [latex]\,f\left(x\right)=-5{x}^{2}+9x-1.[/latex]

As with any quadratic function, the domain is all real numbers.

Because [latex]\,a\,[/latex] is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the [latex]\,x\text{-}[/latex] value of the vertex.

[latex]$$\begin{array}{ccc}\hfill h& =& -\frac{b}{2a}\hfill \\ \hfill & =& -\frac{9}{2\left(-5\right)}\hfill \\ \hfill & =& \frac{9}{10}\hfill \end{array}$$[/latex]

The maximum value is given by [latex]$$\,f\left(h\right).$$[/latex]

[latex]$$\begin{array}{ccc}\hfill f\left(\frac{9}{10}\right)& =& -5{\left(\frac{9}{10}\right)}^{2}+9\left(\frac{9}{10}\right)-1\hfill \\ & =& \frac{61}{20}\hfill \end{array}$$[/latex]

The range is [latex]\,f\left(x\right)\le \frac{61}{20},\,[/latex] or [latex]\,\left(-\infty ,\frac{61}{20}\right].[/latex]

Try It

Find the domain and range of [latex]\,f\left(x\right)=2{\left(x-\frac{4}{7}\right)}^{2}+\frac{8}{11}.[/latex]

Show answer

The domain is all real numbers. The range is [latex]\,f\left(x\right)\ge \frac{8}{11},\,[/latex] or [latex]\,\left[\frac{8}{11},\infty \right).[/latex]

5.7.2 – Determining the Maximum and Minimum Values of Quadratic Functions

The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. We can see the maximum and minimum values in (Figure).

Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).
Figure 9.

There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.

Example 2 – Finding the Maximum Value of a Quadratic Function

A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.

  1. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]\,L.[/latex]
  2. What dimensions should she make her garden to maximize the enclosed area?

Let’s use a diagram such as (Figure) to record the given information. It is also helpful to introduce a temporary variable, [latex]\,W,\,[/latex] to represent the width of the garden and the length of the fence section parallel to the backyard fence.

Diagram of the garden and the backyard.
Figure 10.
  1. We know we have only 80 feet of fence available, and [latex]\,L+W+L=80,\,[/latex] or more simply, [latex]\,2L+W=80.\,[/latex] This allows us to represent the width, [latex]\,W,\,[/latex] in terms of [latex]\,L.[/latex]
    [latex]$$W=80-2L$$[/latex]

    Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so

    [latex]$$\begin{array}{ccc}\hfill A& =& LW=L\left(80-2L\right)\hfill \\ \hfill A\left(L\right)& =& 80L-2{L}^{2}\hfill \end{array}$$[/latex]

    This formula represents the area of the fence in terms of the variable length [latex]\,L.\,[/latex] The function, written in general form, is

    [latex]$$A\left(L\right)=-2{L}^{2}+80L.$$[/latex]
  2. The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since [latex]\,a\,[/latex] is the coefficient of the squared term, [latex]\,a=-2,b=80,\,[/latex] and [latex]\,c=0.[/latex]

To find the vertex:

[latex]$$\begin{array}{ccccccc}\hfill h& =& -\frac{b}{2a}\hfill & & \hfill \phantom{\rule{1em}{0ex}}k& =& A\left(20\right)\hfill \\ & =& -\frac{80}{2\left(-2\right)}\hfill & \phantom{\rule{1em}{0ex}}\text{and}& & =& 80\left(20\right)-2{\left(20\right)}^{2}\hfill \\ & =& 20\hfill & & & =& 800\hfill \end{array}$$[/latex]

The maximum value of the function is an area of 800 square feet, which occurs when [latex]\,L=20\,[/latex] feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.

 

Analysis

This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in (Figure).

Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).
Figure 11.

How To

Given an application involving revenue, use a quadratic equation to find the maximum.

  1. Write a quadratic equation for a revenue function.
  2. Find the vertex of the quadratic equation.
  3. Determine the y-value of the vertex.

Example 3 – Finding Maximum Revenue

The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, [latex]\,p\,[/latex] for price per subscription and [latex]\,Q\,[/latex] for quantity, giving us the equation [latex]\,\text{Revenue}=pQ.[/latex]

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently [latex]\,p=30\,[/latex] and [latex]\,Q=84,000.\,[/latex] We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, [latex]\,p=32\,[/latex] and [latex]\,Q=79,000.\,[/latex] From this we can find a linear equation relating the two quantities. The slope will be

[latex]$$\begin{array}{ccc}\hfill m& =& \frac{79,000-84,000}{32-30}\hfill \\ & =& \frac{-5,000}{2}\hfill \\ & =& -2,500\hfill \end{array}$$[/latex]

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept.

[latex]$$\begin{array}{cccc}\hfill Q& =& -2500p+b\hfill & \phantom{\rule{2em}{0ex}}\text{Substitute in the point}Q=84,000\text{ and }p=30\hfill \\ \hfill 84,000& =& -2500\left(30\right)+b\hfill & \phantom{\rule{2em}{0ex}}\text{Solve for}b\hfill \\ \hfill b& =& 159,000\hfill & \end{array}$$[/latex]

This gives us the linear equation [latex]\,Q=-2,500p+159,000\,[/latex] relating cost and subscribers. We now return to our revenue equation.

[latex]$$\begin{array}{ccc}\hfill \mathrm{Revenue}& =& pQ\hfill \\ \hfill \mathrm{Revenue}& =& p\left(-2,500p+159,000\right)\hfill \\ \hfill \mathrm{Revenue}& =& -2,500{p}^{2}+159,000p\hfill \end{array}$$[/latex]

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

[latex]$$\begin{array}{ccc}\hfill h& =& -\frac{159,000}{2\left(-2,500\right)}\hfill \\ & =& 31.8\hfill \end{array}$$[/latex]

The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

[latex]$$\begin{array}{ccc}\hfill \text{maximum revenue}& =& -2,500{\left(31.8\right)}^{2}+159,000\left(31.8\right)\hfill \\ & =& 2,528,100\hfill \end{array}$$[/latex]

Analysis

This could also be solved by graphing the quadratic as in (Figure). We can see the maximum revenue on a graph of the quadratic function.

Graph of the parabolic function which the x-axis is labeled Price (p) and the y-axis is labeled Revenue (💲). The vertex is at (31.80, 258100).
Figure 12.

Access these online resources for additional instruction and practice with quadratic equations.

Key Equations

general form of a quadratic function [latex]f\left(x\right)=a{x}^{2}+bx+c[/latex]
standard form of a quadratic function [latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]

Key Concepts

  • The domain of a quadratic function is all real numbers. The range varies with the function. See Example 1.
  • A quadratic function’s minimum or maximum value is given by the [latex]\,y\text{-}[/latex] value of the vertex.
  • The minimum or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue. See Example 2 and Example 3.

Glossary

axis of symmetry
a vertical line drawn through the vertex of a parabola, that opens up or down, around which the parabola is symmetric; it is defined by [latex]\,x=-\frac{b}{2a}.[/latex]
general form of a quadratic function
the function that describes a parabola, written in the form [latex]\,f\left(x\right)=a{x}^{2}+bx+c[/latex] , where [latex]\,a,b,\,[/latex] and [latex]\,c\,[/latex] are real numbers and [latex]\,a\ne 0.[/latex]
roots
in a given function, the values of [latex]\,x\,[/latex] at which [latex]\,y=0[/latex] , also called zeros
standard form of a quadratic function
the function that describes a parabola, written in the form [latex]\,f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex] , where [latex]\,\left(h,\text{ }k\right)\,[/latex] is the vertex
vertex
the point at which a parabola changes direction, corresponding to the minimum or maximum value of the quadratic function
vertex form of a quadratic function
another name for the standard form of a quadratic function
zeros
in a given function, the values of [latex]\,x\,[/latex] at which [latex]\,y=0[/latex] , also called roots

License

Icon for the Creative Commons Attribution 4.0 International License

Algebra and Trigonometry Copyright © 2015 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Share This Book