# How do you find the vertex and intercepts for y = -2(x -1)^2 + 3?

May 7, 2018

$\text{see explanation}$

#### Explanation:

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

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

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

$y = - 2 {\left(x - 1\right)}^{2} + 3 \text{ is in vertex form}$

$\text{with "h=1" and } k = 3$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(1 , 3\right)$

$\text{to find the intercepts}$

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

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

$x = 0 \Rightarrow y = - 2 {\left(- 1\right)}^{2} + 3 = 1 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \Rightarrow - 2 {\left(x - 1\right)}^{2} + 3 = 0$

$\text{subtract 3 from both sides}$

$\Rightarrow - 2 {\left(x - 1\right)}^{2} = - 3$

$\text{divide both sides by } - 2$

$\Rightarrow {\left(x - 1\right)}^{2} = \frac{3}{2}$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(x - 1\right)}^{2}} = \pm \sqrt{\frac{3}{2}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x - 1 = \pm \frac{\sqrt{3}}{\sqrt{2}} = \pm \frac{1}{2} \sqrt{6}$

$\text{add 1 to both sides}$

$\Rightarrow x = 1 \pm \frac{1}{2} \sqrt{6} \leftarrow \textcolor{red}{\text{exact solutions}}$

$x \approx - 0.22 \text{ to 2 dec. places ",x~~2.22larrcolor(red)"x-intercepts}$