Chapter 2.1, Problem 81E

To determine

To prove: the x -coordinate of the vertex of the graph is the average of the zeros of

  $f\left(x\right)=a{x}^2+bx+c$ , $a\ne 0$ .

Explanation of Solution

Concept Used:

For the function $f\left(x\right)=a{x}^2+bx+c$ with vertex $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$ ,

1. When $a>0$ , f has a minimum at $x=-\frac{b}{2a}$ and the minimum value is $f\left(-\frac{b}{2a}\right)$ .
2. When $a<0$ , f has a maximum at $x=-\frac{b}{2a}$ and the maximum value is $f\left(-\frac{b}{2a}\right)$ .

From the above definition it is clear that the x -coordinate of the vertex of the graph is equal to $-\frac{b}{2a}$ .

Now to show the x coordinate of the vertex is equal to average of the zeros of $f\left(x\right)=a{x}^2+bx+c$

Using quadratic equation to find zeros of the equation $a{x}^2+bx+c=0$ ,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

SO, the two zeros of the equation $a{x}^2+bx+c=0$ are:

  $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ And $x_2=\frac{-b\sqrt{b^2-4ac}}{2a}$

Now find average of these zeros.

  $\begin{array}{c}\frac{x_1+x_2}{2}=\frac{\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}}{2}\\ =\frac{-b+\sqrt{b^2-4ac}-b-\sqrt{b^2-4ac}}{4a}\\ =-\frac{2b}{4a}\\ =-\frac{b}{2a}\end{array}$

Hence proved that, the x -coordinate of the vertex of the graph is the average of the zeros of

  $f\left(x\right)=a{x}^2+bx+c$ , $a\ne 0$ .