Chapter 11.8, Problem 11MRE

To determine

To find: The coordinates of the vertex and the equation of the axis of symmetry.

Answer to Problem 11MRE

The coordinate of the vertex is $\left(3,0\right)$ and the equation of the symmetry is $x=3$ .

Explanation of Solution

Given information:

The given equation is,

  $y=x^2-6x+9$

The x-coordinate of the vertex is calculated as,

  $\begin{array}{c}x=-\frac{b\left(\text{coefficent of }x\right)}{2a\left(\text{coefficent of }{x}^2\right)}\\ =-\frac{\left(-6\right)}{2\left(1\right)}\\ =3\end{array}$

The y-coordinate of the vertex is calculated as,

  $\begin{array}{l}y=x^2-6x+9\\ y={\left(3\right)}^2-6\left(3\right)+9\\ y=0\end{array}$

The coordinate of the vertex is $\left(3,0\right)$ .

The equation of axis of the symmetry is calculated as,

  $\begin{array}{c}x=-\frac{b\left(\text{coefficent of }x\right)}{2a\left(\text{coefficent of }{x}^2\right)}\\ =-\frac{\left(-6\right)}{2\left(1\right)}\\ x=3\end{array}$

Thus, the coordinate of the vertex is $\left(3,0\right)$ and the equation of the symmetry is $x=3$ .