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

Nov 13, 2017

Vetex is at $\left(3 , 7\right)$ and axis of symmetry is  x = 3 ;

#### Explanation:

$y = - {x}^{2} + 6 x - 2 \mathmr{and} y = - \left({x}^{2} - 6 x\right) - 2$ or

$y = - \left({x}^{2} - 6 x + {3}^{2}\right) + 9 - 2$ or

$y = - {\left(x - 3\right)}^{2} + 7$ . This is vertex form of equation

y=a(x-h)^2+k ; (h,k) being vertex , here $h = 3 , k = 7$

Therefore vetex is at $\left(h , k\right) \mathmr{and} \left(3 , 7\right)$

Axis of symmetry is x= h or x = 3 ;

graph{-x^2+6x-2 [-20, 20, -10, 10]} [Ans]

Nov 13, 2017

$x = 3 \text{ and } \left(3 , 7\right)$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

• " if "a>0" then graph opens up"

• " if "a<0" then graph opens down"

$\text{express y in vertex form using the method of "color(blue)"completing the square}$

• " coefficient of "x^2" term must be 1"

$\Rightarrow y = - 1 \left({x}^{2} - 6 x + 2\right)$

• " add/subtract "(1/2"coefficient of x-term")^2" to "x^2-6x

$\Rightarrow y = - \left({x}^{2} - 6 x \textcolor{red}{+ 9} \textcolor{red}{- 9} + 2\right)$

$\textcolor{w h i t e}{\Rightarrow y} = - {\left(x - 3\right)}^{2} + 7 \leftarrow \textcolor{red}{\text{in vertex form}}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(3 , 7\right)$

$\text{since "a<0" then parabola is vertical and opens down}$

$\text{the axis of symmetry is vertical and passes through the}$
$\text{vertex with equation } x = 3$
graph{(y+x^2-6x+2)(y-1000x+3000)((x-3)^2+(y-7)^2-0.05)=0 [-20, 20, -10, 10]}