Chapter 7, Problem 15CR

To determine

Tofind: thequadratic equation.

Answer to Problem 15CR

The quadratic equation is $f\left(x\right)=\frac{1}{2}\left(x^2+2x-3\right)$ .

Explanation of Solution

Given information:

Minimum value $-2$ and $x$ -intercepts $\text{1 and }-\text{3}$ .

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 $2$ ,

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

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

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

Thus, the equation is $f\left(x\right)=\frac{1}{2}\left(x^2+2x-3\right)$ .