# What is the axis of symmetry and vertex for the graph y=-x^2+6x-4?

Aug 15, 2017

$x = 3 , \left(3 , 5\right)$

#### Explanation:

$\text{given the equation of a parabola in standard form}$

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

$\text{the x-coordinate of the vertex and axis of symmetry is}$

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

$y = - {x}^{2} + 6 x - 4 \text{ is in standard form}$

$\text{with } a = - 1 , b = 6 , c = - 4$

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

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

$\Rightarrow {y}_{\textcolor{red}{\text{vertex}}} = - 9 + 18 - 4 = 5$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(3 , 5\right)$

$\text{equation of axis of symmetry is } x = 3$
graph{(y+x^2-6x+4)(y-1000x+3000)=0 [-10, 10, -5, 5]}