# How do you find the axis of symmetry and vertex point of the function:  y = -x^2 + 6x + 2?

Jan 26, 2018

$x = 3 , \text{ vertex } = \left(3 , 11\right)$

#### Explanation:

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

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

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

${x}_{\textcolor{red}{\text{vertex}}} = - \frac{b}{2 a}$

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

$\text{with "a=-1,b=6" and } c = 2$

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

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

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

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(3 , 11\right)$

$\text{and axis of symmetry is } x = 3$