# How do you find the vertex and intercepts for y = 2(x – 5)^2 + 2?

Apr 29, 2017

$\left(5 , 2\right) , \left(0 , 52\right) \text{ no x-intercepts}$

#### Explanation:

The equation of a parabola in $\textcolor{b l u e}{\text{vertex form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where (h , k) are the coordinates of the vertex and a is a constant.

$y = 2 {\left(x - 5\right)}^{2} + 2 \text{ is in this form}$

$\text{with " (h,k)=(5,2)larrcolor(red)" vertex}$

$\textcolor{b l u e}{\text{ to find intercepts}}$

• " let x=0, in equation, for y-intercept"

• " let y=0, in equation, for x-intercept"

$x = 0 \to y = 2 {\left(- 5\right)}^{2} + 2 = 52 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to 2 {\left(x - 5\right)}^{2} + 2 = 0$

$\Rightarrow {\left(x - 5\right)}^{2} = \frac{- 2}{2} = - 1$

${\left(x - 5\right)}^{2} = - 1 \text{ has no real solutions thus no x-intercepts}$
graph{2(x-5)^2+2 [-10, 10, -5, 5]}