# 5.1 Quadratic functions  (Page 4/15)

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## Finding the domain and range of a quadratic function

Find the domain and range of $\text{\hspace{0.17em}}f\left(x\right)=-5{x}^{2}+9x-1.$

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

Because $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the $\text{\hspace{0.17em}}x\text{-}$ value of the vertex.

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

The maximum value is given by $\text{\hspace{0.17em}}f\left(h\right).$

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

The range is $\text{\hspace{0.17em}}f\left(x\right)\le \frac{61}{20},\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\left(-\infty ,\frac{61}{20}\right].$

Find the domain and range of $\text{\hspace{0.17em}}f\left(x\right)=2{\left(x-\frac{4}{7}\right)}^{2}+\frac{8}{11}.$

The domain is all real numbers. The range is $\text{\hspace{0.17em}}f\left(x\right)\ge \frac{8}{11},\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\left[\frac{8}{11},\infty \right).$

## 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 [link] .

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.

## 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 $\text{\hspace{0.17em}}L.$
2. What dimensions should she make her garden to maximize the enclosed area?

Let’s use a diagram such as [link] to record the given information. It is also helpful to introduce a temporary variable, $\text{\hspace{0.17em}}W,\text{\hspace{0.17em}}$ to represent the width of the garden and the length of the fence section parallel to the backyard fence.

1. We know we have only 80 feet of fence available, and $\text{\hspace{0.17em}}L+W+L=80,\text{\hspace{0.17em}}$ or more simply, $\text{\hspace{0.17em}}2L+W=80.\text{\hspace{0.17em}}$ This allows us to represent the width, $\text{\hspace{0.17em}}W,\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}L.$
$W=80-2L$

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

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

This formula represents the area of the fence in terms of the variable length $\text{\hspace{0.17em}}L.\text{\hspace{0.17em}}$ The function, written in general form, is

$A\left(L\right)=-2{L}^{2}+80L.$
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 $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is the coefficient of the squared term, $\text{\hspace{0.17em}}a=-2,b=80,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}c=0.$

To find the vertex:

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

The maximum value of the function is an area of 800 square feet, which occurs when $\text{\hspace{0.17em}}L=20\text{\hspace{0.17em}}$ 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.

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.

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