Chapter 1.2, Problem 38E

Solution

To find: The minimum or maximum value of a quadratic function.

The minimum value of the function $y=-12x\left(x-9\right)$ is 243 that occurs at $x=4.5$ .

Given data: The given quadratic function is $y=-12x\left(x-9\right)$ .

Method/Formula used:

The form of a quadratic function $y=f\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)$ is known as intercept form that represents a parabola with vertex is $\left(h,k\right)$ where $h=\frac{x_1+x_2}{2}$ and $k=f\left(h\right)$ or $k=a\left(h-x_1\right)\left(h-x_2\right)$ .The maximum or minimum of the function $f\left(x\right)$ occurs at the vertex at the vertex $\left(h,k\right)$ . If $a>0$ then k is minimum value of the function $f\left(x\right)$ , and if $a<0$ then k is maximum value of the function $y\left(x\right)$ .

Calculation:

The given function is $y\left(x\right)=-12\left(x-0\right)\left(x-9\right)$ . Comparing it with the standard equation $y=a\left(x-x_1\right)\left(x-x_2\right)$ , $a=-12<0$ , $x_1=0$ , and $x_2=9$ .

Therefore, the vertex $\left(h,k\right)$ of the graph of the function $y\left(x\right)$ is

  $\begin{array}{c}h=\left(x_1+x_2\right)/2\\ =\left(0+9\right)/2\\ =9/2\\ =4.5\end{array}$

  $\begin{array}{c}k=y\left(4.5\right)\\ =-12\left(4.5\right)\left(4.5-9\right)\text{}\text{}\text{}\text{}\leftarrow \text{Substitute}x=4.5\text{in}y(x)\\ =-12\left(4.5\right)\left(-4.5\right)\\ =243\end{array}$

Since, $a=-12<0$ , therefore $k=243$ is the maximum value of the function $y=-12x\left(x-9\right)$ , that occurs for $x=4.5$ .