# How do you find the axis of symmetry and the vertex for the line y = x^2 - 6x +11?

May 30, 2018

$x = 3 \text{ and } \left(3 , 2\right)$

#### Explanation:

$\text{given the equation of a quadratic in "color(blue)"standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$\text{then the x-coordinate of the vertex which is also the}$
$\text{equation of the axis of symmetry is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$y = {x}^{2} - 6 x + 11 \text{ is in standard form}$

$\text{with "a=1,b=-6" and } c = 11$

${x}_{\text{vertex}} = - \frac{- 6}{2} = 3$

$\text{substitute this value into the equation for y}$

${y}_{\text{vertex}} = {3}^{2} - \left(6 \times 3\right) + 11 = 2$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(3 , 2\right)$

$\text{axis of symmetry is } x = 3$
graph{(y-x^2+6x-11)(y-1000x+3000)=0 [-10, 10, -5, 5]}