3.2 Quadratic functions  (Page 4/14)

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

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

The maximum value is given by $f(h).$

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

Let’s use a diagram such as [link] to record the given information. It is also helpful to introduce a temporary variable, $W,$ 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 $L+W+L=80,$ or more simply, $2L+W=80.$ This allows us to represent the width, $W,$ in terms of $L.$
   $$W=80-2L$$

   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

   $$\begin{array}{l}A=LW=L(80-2L)\hfill \\ A(L)=80L-2L^2\hfill \end{array}$$

   This formula represents the area of the fence in terms of the variable length $L.$ The function, written in general form, is

   $$A(L)=-2L^2+80L.$$
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 $a$ is the coefficient of the squared term, $a=\mathrm{-2},b=80,$ and $c=0.$

To find the vertex:

The maximum value of the function is an area of 800 square feet, which occurs when $L=20$ 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.