Chapter 7.7, Problem 36WE

To determine

Tofind: thequadratic equation.

Answer to Problem 36WE

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

Explanation of Solution

Given information:

Range: $\left\{y:y\le 9\right\}$ ; x -intercept $-2\text{ and 4}$ .

Calculation:

to find the quadratic equation, first use the formula of quadratic formula with roots,

$f\left(x\right)=a\left[x^2-\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)\right]$

Now, substitute and simplify,

$\begin{array}{l}f\left(x\right)=a\left[x^2-\left(-2+4\right)x+\left(\left(-2\right)\left(4\right)\right)\right]\text{ [substitute]}\\ f\left(x\right)=a\left(x^2-2x-8\right)\text{ [simplify]}\end{array}$

Since, Range: $\left\{y:y\le 9\right\}$

Thus, the maximum value of equation is $9$ which occur at axis,

Now, to find the axis add both x- intercepts and divide by $2$ ,

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

Now, substitute $\left(1,9\right)$ coordinates in the equation,

$\begin{array}{l}f\left(x\right)=a\left(x^2-2x-8\right)\\ 9=a\left({\left(1\right)}^2-2\left(1\right)-8\right)\text{ [substitute]}\\ 9=a\left(-9\right)\text{ [add and subtract]}\\ a=-1\text{ [simplify]}\end{array}$

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