# How do you find the vertex and intercepts for y=2(x-3)^2 +4?

Apr 10, 2017

see explanation.

#### 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.

$y = 2 {\left(x - 3\right)}^{2} + 4 \text{ is in this form}$

$\text{with " a=2, h=3" and } k = 4$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(3 , 4\right)$

$\text{since " a>0" then min. turning point } \bigcup$

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

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

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

$x = 0 \to y = 2 {\left(- 3\right)}^{2} + 4 = 22 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to 2 {\left(x - 3\right)}^{2} + 4 = 0$

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

This has no real solutions hence f(x) has no x-intercepts.
graph{2(x-3)^2+4 [-31.56, 31.67, -15.8, 15.8]}