Chapter 7.7, Problem 28WE

To determine

Tofind: thequadratic equation.

Answer to Problem 28WE

The quadratic equation is $f\left(x\right)=\frac{4}{5}\left(x^2-x-6\right)$ .

Explanation of Solution

Given information:

Minimum value $-5$ and $x$ -intercepts $-2\text{ and 3}$ .

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 $2$ ,

$\begin{array}{l}x=\frac{-2+3}{2}\\ x=\frac{1}{2}\text{ [simplify]}\end{array}$

Therefore vertex is $\left(\frac{1}{2},-5\right)$ . Since, the vertex lies on the graph of quadratic equation, so substitute its coordinates in the equation,

$\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\\ f\left(x\right)=a\left(x+2\right)\left(x-3\right)\text{ [because }x-\text{intercept of equation is 0}-2,3]\\ f\left(x\right)=a\left(x^2-x-6\right)\text{ [multiply]}\\ -5=a\left({\left(\frac{1}{2}\right)}^2-\left(\frac{1}{2}\right)-6\right)\text{ [substitute]}\\ -5=a\left(-\frac{25}{4}\right)\text{ [add and subtract]}\\ a=\frac{4}{5}\text{ [simplify]}\end{array}$

Thus, the equation is $f\left(x\right)=\frac{4}{5}\left(x^2-x-6\right)$ .