Chapter 9.3, Problem 6CYU

To determine

To solve: The equation by examining the given graph of the related function.

Answer to Problem 6CYU

The equation of the graph is $\frac{1}{4}{x}^2-2x+3$ .

Explanation of Solution

Given information: The figure of graph is given.

Calculation: : The figure of graph is given.

The equation of intercept form of parabola is $y=a(x-m)(x-n)$ .

The graph has $x-$ intercept at $(2,0)\mathrm{and}(6,0)$ .

So put $m=2\mathrm{and}n=6$ in intercept form of parabola.

  $\begin{array}{l}y=a(x-m)(x-n)\\ y=a(x-2)(x-6)\end{array}$

Parabola passes through the points $(0,3)$ so put $x=0\mathrm{and}y=3\mathrm{in}y=a(x-2)(x-6)$

  $\begin{array}{l}y=a(x-2)(x-6)\\ 3=a(0-2)(0-6)\\ 3=a(-2)(-6)\\ 3=12a\\ \frac{3}{12}=a\\ \frac{1}{4}=a\end{array}$

Put $a=\frac{1}{4}\mathrm{in}y=a(x-2)(x-6)$ then:

  $\begin{array}{l}y=a(x-2)(x-6)\\ y=\frac{1}{4}(x^2-6x-2x+12)\\ y=\frac{1}{4}(x^2-8x+12)\\ y=\frac{1}{4}{x}^2-\frac{8}{4}x+\frac{12}{4}\\ y=\frac{1}{4}{x}^2-2x+3\end{array}$

Thus, the equation of graph is $\frac{1}{4}{x}^2-2x+3$ .