Chapter 2.2, Problem 74E

a.

To determine

To calculate: What does $f\left(\frac{p+q}{2}\right)$ represent in the graph.

Answer to Problem 74E

  $f\left(\frac{p+q}{2}\right)$ represents the maximum height of the graph.

Explanation of Solution

Given Information: The graph is represented by the equation $f\left(x\right)=a\left(x-p\right)\left(x-q\right)$

Calculation:

Simplify the equation,

  $\begin{array}{c}f\left(x\right)=a\left(x-p\right)\left(x-q\right)\\ =a\left(x^2-\left(p+q\right)x+pq\right)\\ =a{x}^2-a\left(p+q\right)x+apq\end{array}$

The x -coordinate of the vertex of the parabola of the form $y=a{x}^2+bx+c$ is given by $x=-\frac{b}{2a}$ and y- coordinate is $f\left(x\right)$ .

The x- coordinate of the parabola $f\left(x\right)=a{x}^2-a\left(p+q\right)x+apq$ is,

  $\begin{array}{c}x=-\frac{b}{2a}\\ =-\frac{-a\left(p+q\right)}{2a}\left(a=a,b=-a\left(p+q\right)\right)\\ =\frac{p+q}{2}\end{array}$

The y -coordinate is,

  $\begin{array}{c}f\left(x\right)=a{x}^2-a\left(p+q\right)x+apq\\ f\left(\frac{p+q}{2}\right)=a{\left(\frac{p+q}{2}\right)}^2-a\left(p+q\right)\left(\frac{p+q}{2}\right)+apq\\ =apq\end{array}$

Therefore, the vertex is $\left(\left(\frac{p+q}{2}\right),f\left(\frac{p+q}{2}\right)\right)=\left(\left(\frac{p+q}{2}\right),apq\right)$ .

Since $a>0$ , $f\left(\frac{p+q}{2}\right)$ represents the maximum height of the graph.

b.

To determine

To calculate: What does $f\left(\frac{p+q}{2}\right)$ represent in the graph, if $a<0$ .

Answer to Problem 74E

  $f\left(\frac{p+q}{2}\right)$ represents the minimum height of the graph.

Explanation of Solution

Given Information: The graph is represented by the equation $f\left(x\right)=a\left(x-p\right)\left(x-q\right),a<0$

Calculation:

Simplify the equation,

  $\begin{array}{c}f\left(x\right)=a\left(x-p\right)\left(x-q\right)\\ =a\left(x^2-\left(p+q\right)x+pq\right)\\ =a{x}^2-a\left(p+q\right)x+apq\end{array}$

The x -coordinate of the vertex of the parabola of the form $y=a{x}^2+bx+c$ is given by $x=-\frac{b}{2a}$ and y- coordinate is $f\left(x\right)$ .

The x- coordinate of the parabola $f\left(x\right)=a{x}^2-a\left(p+q\right)x+apq$ is,

  $\begin{array}{c}x=-\frac{b}{2a}\\ =-\frac{-a\left(p+q\right)}{2a}\left(a=a,b=-a\left(p+q\right)\right)\\ =\frac{p+q}{2}\end{array}$

The y -coordinate is,

  $\begin{array}{c}f\left(x\right)=a{x}^2-a\left(p+q\right)x+apq\\ f\left(\frac{p+q}{2}\right)=a{\left(\frac{p+q}{2}\right)}^2-a\left(p+q\right)\left(\frac{p+q}{2}\right)+apq\\ =apq\end{array}$

Therefore, the vertex is $\left(\left(\frac{p+q}{2}\right),f\left(\frac{p+q}{2}\right)\right)=\left(\left(\frac{p+q}{2}\right),apq\right)$ .

Since $a<0$ , $f\left(\frac{p+q}{2}\right)$ represents the minimum height of the graph.