Chapter 1.2, Problem 36E

Solution

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

The minimum value of the function $f\left(x\right)=3\left(x+10\right)\left(x-8\right)$ is -243 that occurs at $x=-1$ .

Given data: The given quadratic function is $f\left(x\right)=3\left(x+10\right)\left(x-8\right)$ .

Method/Formula used:

The form of a quadratic function $f\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)$ 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 $f\left(x\right)=3\left(x+10\right)\left(x-8\right)$ . Comparing it with the standard equation $f\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)$ , $a=3>0$ , $x_1=-10$ , and $x_2=8$ .

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

  $\begin{array}{c}h=\left(x_1+x_2\right)/2\\ =\left(-10+8\right)/2\\ =-2/2\\ =-1\end{array}$

  $\begin{array}{c}k=f\left(-1\right)\\ =3\left(-1+10\right)\left(-1-8\right)\\ =3\left(9\right)\left(-9\right)\\ =-243\end{array}$

Since, $a>0$ , therefore $k=-243$ is the minimum value of the function $f\left(x\right)=3\left(x+10\right)\left(x-8\right)$ , that occurs for $x=-1$ .