# What is the vertex of y=-2x^2 + 2x + 5 ?

Aug 10, 2017

$\left(\frac{1}{2} , \frac{11}{2}\right)$

#### Explanation:

$\text{given the equation of a parabola in standard form}$

$\text{that is } y = a {x}^{2} + b x + c$

"then "x_(color(red)"vertex")=-b/(2a)

$y = - 2 {x}^{2} + 2 x + 5 \text{ is in standard form}$

$\text{with } a = - 2 , b = + 2 , c = 5$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{2}{- 4} = \frac{1}{2}$

$\text{substitute this value into the equation for the corresponding}$
$\text{y-coordinate}$

<$\Rightarrow {y}_{\textcolor{red}{\text{vertex}}} = - 2 {\left(\frac{1}{2}\right)}^{2} + 2 \left(\frac{1}{2}\right) + 5 = \frac{11}{2}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{1}{2} , \frac{11}{2}\right)$

Aug 10, 2017

Vertex is at $\left(\frac{1}{2} , \frac{11}{2}\right)$.

#### Explanation:

The axis of symmetry is also the x value of the vertex. So we can use the formula $x = \frac{- b}{2 a}$ to find the axis of symmetry.

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

$x = \frac{1}{2}$

Substitute $x = \frac{1}{2}$ back into the original equation for the y value.

$y = - 2 {\left(\frac{1}{2}\right)}^{2} + 2 \left(\frac{1}{2}\right) + 5$

$y = \frac{11}{2}$

Therefore, the vertex is at $\left(\frac{1}{2} , \frac{11}{2}\right)$.