# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function y-2=2(x-3)^2?

Mar 19, 2017

see explanation.

#### Explanation:

$\text{ Express " y-2=2(x-3)^2" in the form}$

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

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.

$y = 2 {\left(x - 3\right)}^{2} + 2 \text{ is in this form}$

$\text{here " h=3" and } k = 2$

$\Rightarrow \text{ vertex } = \left(3 , 2\right)$

To determine min/max consider the value of a

• a>0rArr" minimum " uuu

• a<0rArr" maximum " nnn

$\text{here " a=2rArr" minimum}$

The axis of symmetry passes through the vertex and is vertical with equation $\textcolor{b l u e}{x = 3}$

The minimum value at the vertex is y = 2

$\textcolor{b l u e}{\text{Intercepts}}$

$x = 0 \to y = 2 {\left(- 3\right)}^{2} + 2 = 20 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to 2 {\left(x - 3\right)}^{2} = - 2$

$\Rightarrow {\left(x - 3\right)}^{2} = - 1$ which has no real solutions and therefore graph does not cross the x-axis.
graph{(y-2x^2+12x-20)(y-1000x+3000)=0 [-40, 40, -20, 20]}