Chapter 7.7, Problem 27WE

To determine

Tofind: thequadratic equation.

Answer to Problem 27WE

The quadratic equation is $f\left(x\right)=\frac{1}{4}\left(x^2-8x\right)$ .

Explanation of Solution

Given information:

Minimum value $-8$ and $x$ -intercepts $\text{0 and 8}$ .

Calculation: to find the quadratic equation, first find the axis of the quadratic equation graph, because the min value occur at the axis,

To find the axis, add the $x$ -intercepts an divide by $2$ ,

$\begin{array}{l}x=\frac{0+8}{2}\\ x=4\text{ [simplify]}\end{array}$

Therefore vertex is $\left(4,-8\right)$ . Since, the vertex lies on the graph of quadratic equation, so substitute its coordinates in the equation,

$\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\\ f\left(x\right)=a\left(x-0\right)\left(x-8\right)\text{ [because }x-\text{intercept of equation is 0},8]\\ f\left(x\right)=a\left(x^2-8x\right)\text{ [multiply]}\\ -8=a\left({\left(4\right)}^2-8\left(4\right)\right)\text{ [substitute]}\\ -8=a\left(-32\right)\text{ [add and subtract]}\\ a=\frac{1}{4}\text{ [simplify]}\end{array}$

Thus, the equation is $f\left(x\right)=\frac{1}{4}\left(x^2-8x\right)$ .