# How do you find the vertex and intercepts for y+12x-2x^2=15?

Dec 12, 2017

$\text{see explanation}$

#### Explanation:

$\text{express in standard form}$

$\Rightarrow y = 2 {x}^{2} - 12 x + 15$

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

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

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

$\text{to obtain this form use the method of "color(blue)"completing the square}$

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

$\Rightarrow 2 \left({x}^{2} - 6 x\right) + 15$

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

$2 \left({x}^{2} + 2 \left(- 3\right) x \textcolor{red}{+ 9} \textcolor{red}{- 9}\right) + 15$

$= 2 {\left(x - 3\right)}^{2} - 18 + 15$

$\Rightarrow y = 2 {\left(x - 3\right)}^{2} - 3 \leftarrow \textcolor{red}{\text{in vertex form}}$

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

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

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

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

$x = 0 \to y = 2 {\left(- 3\right)}^{2} - 3 = 15 \leftarrow \textcolor{red}{\text{y-intercept}}$

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

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

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

$\Rightarrow x - 3 = \pm \sqrt{\frac{3}{2}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x = 3 \pm \sqrt{\frac{3}{2}} \leftarrow \textcolor{red}{\text{x-intercepts}}$

$\Rightarrow x \approx 1.78 , x \approx 4.22 \text{ to 2 dec. places}$