Chapter 7.7, Problem 26WE

To determine

Tofind: thequadratic equation.

Answer to Problem 26WE

The quadratic equation is $f\left(x\right)=-\frac{2}{3}\left(x^2-2x-8\right)$ .

Explanation of Solution

Given information:

Maximum value $6$ and $x$ -intercepts $-2\text{ and 4}$ .

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 $2$ ,

$\begin{array}{l}x=\frac{-2+4}{2}\\ x=1\text{ [simplify]}\end{array}$

Therefore vertex is $\left(1,6\right)$ . Since, the vertex lies on the graph of quadratic equation, so substitute its coordinates in the equation,

$\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\\ f\left(x\right)=a\left(x+2\right)\left(x-4\right)\text{ [because }x-\text{intercept of equation is }-2,4]\\ f\left(x\right)=a\left(x^2-2x-8\right)\text{ [multiply]}\\ 6=a\left({\left(1\right)}^2-2\left(1\right)-8\right)\text{ [substitute]}\\ 6=a\left(-9\right)\text{ [add and subtract]}\\ a=-\frac{2}{3}\text{ [simplify]}\end{array}$

Thus, the equation is $f\left(x\right)=-\frac{2}{3}\left(x^2-2x-8\right)$ .