Chapter 7, Problem 13CT

To determine

Tofind: thequadratic equation.

Answer to Problem 13CT

The quadratic equation is 13

  $f\left(x\right)=\frac{9}{4}\left(\frac{4x^2+4x-15}{4}\right)$ .

Explanation of Solution

Given information:

Minimum value $-9$ and zeros $\frac{3}{2}\text{ and }-\frac{5}{2}$ .

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 zerosan divide by $2$ ,

  $\begin{array}{l}x=\frac{\frac{3}{2}-\frac{5}{2}}{2}\\ x=-\frac{1}{2}\text{ [simplify]}\end{array}$

Therefore vertex is $\left(-\frac{1}{2},-9\right)$ . Since, the vertex lies on the graph of quadratic equation, so substitute its coordinates in the equation,

  $\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\\ f\left(x\right)=a\left(x-\frac{3}{2}\right)\left(x+\frac{5}{2}\right)\text{ [because zeros of equation is }\frac{3}{2},-\frac{5}{2}]\\ f\left(x\right)=a\left(\frac{4x^2+4x-15}{4}\right)\text{ [multiply]}\\ -9=a\left(\frac{4{\left(-\frac{1}{2}\right)}^2+4\left(-\frac{1}{2}\right)-15}{4}\right)\text{ [substitute]}\\ -9=a\left(-4\right)\text{ [add and subtract]}\\ a=\frac{9}{4}\text{ [simplify]}\end{array}$

Thus, the equation is $f\left(x\right)=\frac{9}{4}\left(\frac{4x^2+4x-15}{4}\right)$ .