Chapter 7.7, Problem 31WE

To determine

Tofind: thequadratic equation.

Answer to Problem 31WE

The quadratic equation is $f\left(x\right)=\frac{2}{3}\left(x^2-4x-5\right)$ .

Explanation of Solution

Given information:

Minimum value $-6$ and zeros of function $-1\text{ and 5}$ .

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,

$\begin{array}{l}f\left(x\right)=a\left[x^2-\left(-1+5\right)x+\left(\left(-1\right)\left(5\right)\right)\right]\text{ [substitute]}\\ f\left(x\right)=a\left(x^2-4x-5\right)\text{ [simplify]}\end{array}$

To find the axis, add the zeros of function and divide by $2$ ,

$\begin{array}{l}x=\frac{-1+5}{2}\\ x=2\text{ [simplify]}\end{array}$

Therefore vertex is $\left(2,-6\right)$ . Since, the vertex lies on the graph of quadratic equation, so substitute its coordinates in the equation,

$\begin{array}{l}f\left(x\right)=a\left(x^2-4x-5\right)\\ -6=a\left({\left(2\right)}^2-4\left(2\right)-5\right)\text{ [substitute]}\\ -6=a\left(-9\right)\text{ [add and subtract]}\\ a=\frac{2}{3}\text{ [simplify]}\end{array}$

Thus, the equation is $f\left(x\right)=\frac{2}{3}\left(x^2-4x-5\right)$ .