Chapter 2.2, Problem 68E

To determine

To write: Two different quadratic functions in intercept form whose graphs have the axis of symmetry $x=3$ .

Answer to Problem 68E

The two different equations are $y={\left(x-3\right)}^2\text{ and }y={\left(2x-6\right)}^2$ .

Explanation of Solution

Given Information:

The axis of symmetry is $x=3$ .

Calculation:

The axis of symmetry of the quadratic function whose standard form is $y=a{x}^2+bx+c$ is,

  $x=-\frac{b}{2a}$

Here, $x=3$ . Therefore,

  $\begin{array}{c}-\frac{b}{2a}=3\\ -b=6a\\ b=-6a\end{array}$

Simplify the standard from of the quadratic function as,

  $\begin{array}{c}y=a{x}^2+bx+c\\ =a{x}^2+\left(-6a\right)x+c\\ =a{x}^2-6ax+c\end{array}$

Choose $a=1\text{ and }c=9$ to write the first equation,

  $\begin{array}{c}y=a{x}^2-6ax+c\\ =x^2-6x+9\\ ={\left(x-3\right)}^2\end{array}$

Thus, the first quadratic functions in intercept form whose graphs have the axis of symmetry $x=3$ is $y={\left(x-3\right)}^2$ .

Choose $a=4\text{ and }c=36$ to write the first equation,

  $\begin{array}{c}y=a{x}^2-6ax+c\\ =4x^2-24x+36\\ ={\left(2x\right)}^2-2\times 2x\times 6+{\left(6\right)}^2\\ ={\left(2x-6\right)}^2\end{array}$

Thus, the first quadratic functions in intercept form whose graphs have the axis of symmetry $x=3$ is $y={\left(2x-6\right)}^2$ .