Chapter 4, Problem 26MCQ

Solution

The relationship between the $x$ -intercepts of the graph of a quadratic function and the $x$ -coordinate of the vertex of that graph.

It has been determined that the $x$ -coordinate of the vertex of the graph of a quadratic function equals the average of the $x$ -intercepts of that graph.

Given:

A quadratic function, $f\left(x\right)=a{x}^2+bx+c$ .

Concept used:

The vertex of the parabola formed by graphing the quadratic function $f\left(x\right)=a{x}^2+bx+c$ , is given as $\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)$ .

Calculation:

The given quadratic function is, $f\left(x\right)=a{x}^2+bx+c$ .

Using the quadratic formula, the roots of this function are given as:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now, by definition, these roots are nothing but the $x$ -intercepts of the graph of this quadratic function.

So, the $x$ -intercepts of the graph of the quadratic function are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

The average of these $x$ -intercepts is: $\frac{\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}}{2}=\frac{2\left(\frac{-b}{2a}\right)}{2}=\frac{-b}{2a}$ .

As discussed previously, the coordinates of the vertex of the parabola formed by graphing the quadratic function $f\left(x\right)=a{x}^2+bx+c$ , is given as $\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)$ .

Then, the $x$ -coordinate of this vertex is $-\frac{b}{2a}$ .

So, comparing, it follows that the $x$ -coordinate of the vertex of the graph of a quadratic function equals the average of the $x$ -intercepts of that graph.

Conclusion:

It has been determined that the $x$ -coordinate of the vertex of the graph of a quadratic function equals the average of the $x$ -intercepts of that graph.