# What is the vertex of y= -2x^2 + 2x + 9 ?

Jun 13, 2018

$\text{vertex } = \left(\frac{1}{2} , \frac{19}{2}\right)$

#### Explanation:

$\text{given a quadratic in standard form}$

y=ax^2+bx+c;a!=0

$\text{then the x-coordinate of the vertex is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$y = - 2 {x}^{2} + 2 x + 9 \text{ is in standard form}$

$\text{with "a=-2,b=2" and } c = 9$

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

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

${y}_{\text{vertex}} = - 2 {\left(\frac{1}{2}\right)}^{2} + 2 \left(\frac{1}{2}\right) + 9 = \frac{19}{2}$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{1}{2} , \frac{19}{2}\right)$