# How do you find the vertex of the parabola y=-2x^2 + 12x - 13?

Mar 15, 2018

$\text{vertex } = \left(3 , 5\right)$

#### Explanation:

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

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

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

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

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

$\Rightarrow y = - 2 \left({x}^{2} - 6 x + \frac{13}{2}\right)$

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

$y = - 2 \left({x}^{2} + 2 \left(- 3\right) x \textcolor{red}{+ 9} \textcolor{red}{- 9} + \frac{13}{2}\right)$

$\textcolor{w h i t e}{y} = - 2 {\left(x - 3\right)}^{2} - 2 \left(- 9 + \frac{13}{2}\right)$

$\textcolor{w h i t e}{y} = - 2 {\left(x - 3\right)}^{2} + 5 \leftarrow \textcolor{red}{\text{in vertex form}}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(3 , 5\right)$
graph{(y+2x^2-12x+13)((x-3)^2+(y-5)^2-0.04)=0 [-10, 10, -5, 5]}

Mar 15, 2018

Vertex is at $\left(3 , 5\right)$ .

#### Explanation:

$y = - 2 {x}^{2} + 12 x - 13 \mathmr{and} y = - 2 \left({x}^{2} - 6 x\right) - 13$ or

$y = - 2 \left({x}^{2} - 6 x + 9\right) + 18 - 13$ or

$f \left(x\right) = - 2 {\left(x - 3\right)}^{2} + 5$. Comparing with vertex form of

equation y = a(x-h)^2+k ; (h,k) being vertex we find

here $h = 3 , k = 5 \therefore$ Vertex is at $\left(3 , 5\right)$ .

graph{-2x^2+12x-13 [-17.78, 17.78, -8.89, 8.89]} [Ans]