Chapter 1.2, Problem 50E

Solution

To show: The vertices of quadratic functions $y=a{\left(x-h\right)}^2+k$ and $y=a\left(x-p\right)\left(x-q\right)$ occur at $x=h$ , $x=\frac{p+q}{2}$ respectively..

Given data: The given form of functions are $y=a{\left(x-h\right)}^2+k$ and $y=a\left(x-p\right)\left(x-q\right)$ .

Method/Formula used:

The form of quadratic function $y=a{x}^2+bx+c$ is known as the standard form of a quadratic function. The graph of a quadratic function is a parabola.

Formulae:

(1) ${\left(a-b\right)}^2=a^2-2a\cdot b+b^2$

(2) $\left(a+b\right)\left(c+d\right)=a\cdot c+a\cdot d+b\cdot c+b\cdot d$

FOIL Method of Multiplication.

(3) $m\left(a+b\right)=m\cdot a+m\cdot b$

Distributive Property of Multiplication

Proof:

The given functions are $y=a{\left(x-h\right)}^2+k$ and $y=a\left(x-p\right)\left(x-q\right)$ .

Standard form of the function $y=a{\left(x-h\right)}^2+k$ can be obtained as:

  $\begin{array}{c}y=a{\left(x-h\right)}^2+k\text{}\text{}\text{}\text{}\text{}\leftarrow \text{Original}\text{function}\\ =a\left(x^2-2\cdot x\cdot h+h^2\right)+k\leftarrow \text{Use Formula }\text{(1)}\\ =\left(a{x}^2-a\cdot 2\cdot x\cdot h\cdot +a\cdot h^2\right)+k\leftarrow \text{Use distributive property}\\ =\left(a{x}^2-a\cdot 2\cdot x\cdot h\cdot +a\cdot h^2\right)+k\leftarrow \text{simplify}\\ =a{x}^2+\left(-2ah\right)x+\left(a{h}^2+k\right)\text{Standard}\text{Form}\end{array}$

Thus, the standard form of the function $y=a{\left(x-h\right)}^2+k$ is

  $y\left(x\right)=a{x}^2+\left(-2ah\right)x+\left(a{h}^2+k\right)$ ... (1)

Standard form of the function $y=a\left(x-p\right)\left(x-q\right)$ can be obtained as:

  $\begin{array}{c}y=a\left(x-p\right)\left(x-q\right)\leftarrow \text{Original}\text{function}\\ =a\left[x\cdot x+x\cdot \left(-q\right)+\left(-p\right)\cdot x+\left(-p\right)\cdot \left(-q\right)\right]\leftarrow \text{Use FOIL Method}\\ =a\left[x^2-qx-p\cdot x+pq\right]\leftarrow \text{simplify}\\ =\left[a\cdot x^2-a\cdot qx-a\cdot p\cdot x+a\cdot pq\right]\leftarrow \text{Use distributive property}\\ =\left[a{x}^2+\left(-aq-ap\right)x+apq\right]\leftarrow \text{simplify}\\ =a{x}^2+\left(-aq-ap\right)x+apq\text{Standard}\text{form}\end{array}$

Thus, the standard form of the function $y=a\left(x-p\right)\left(x-q\right)$ is

  $y\left(x\right)=a{x}^2+\left(-aq-ap\right)x+apq$ ... (2)

Case I:

The vertex of the graph of the quadratic function $y=a{x}^2+bx+c$ occurs at $x=\frac{-b}{2a}$ .

Comparing the function $y=a{\left(x-h\right)}^2+k\equiv a{x}^2+\left(-2ah\right)x+\left(a{h}^2+k\right)$ with the function $y=a{x}^2+bx+c$ ,

$y=a{x}^2+bx+c$	$y=a{\left(x-h\right)}^2+k\equiv a{x}^2+\left(-2ah\right)x+\left(a{h}^2+k\right)$
Vertex at $x=\frac{-b}{2a}$	Vertex at $x=\frac{-\left(-2ah\right)}{2a}=h$

Thus the vertex of the function $y=a{\left(x-h\right)}^2+k$ occurs at $x=h$ . Hence, proved.

Case II:

The vertex of the graph of the quadratic function $y=a{x}^2+bx+c$ occurs at $x=\frac{-b}{2a}$ .

Comparing the function $y=a\left(x-p\right)\left(x-q\right)\equiv a{x}^2+\left(-aq-ap\right)x+apq$ with the function $y=a{x}^2+bx+c$ ,

$y=a{x}^2+bx+c$	$y=a\left(x-p\right)\left(x-q\right)\equiv a{x}^2+\left(-aq-ap\right)x+apq$
Vertex at $x=\frac{-b}{2a}$	Vertex at   $x=\frac{-\left(-ap-aq\right)}{2a}=\frac{a\left(p+q\right)}{2a}=\frac{p+q}{2}$

Thus the vertex of the function $y=a\left(x-p\right)\left(x-q\right)$ occurs at $x=\frac{p+q}{2}$ . Hence, proved.