# How do you find the vertex and the intercepts for f(x)=-2x^2+2x+4?

Nov 25, 2017

Vertex $\left(\frac{1}{2} , \frac{9}{2}\right)$
Y- intercept $\left(0 , 4\right)$
X - intercept $\left(- 1 , 0\right)$
X - intercept $\left(2 , 0\right)$

#### Explanation:

Given -

$f \left(x\right) = - 2 {x}^{2} + 2 x + 4$

$y = - 2 {x}^{2} + 2 x + 4$

Vertex

$x = \frac{- b}{2 a} = \frac{- 2}{2 \times \left(- 2\right)} = \frac{- 2}{- 4} = \frac{1}{2}$

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

$y = - 2 \times {\left(\frac{1}{2}\right)}^{2} + \left(2 \times \frac{1}{2}\right) + 4$

$y = - 2 \times \left(\frac{1}{4}\right) + 1 + 4$

$y = - - \frac{1}{2} + 1 + 4 = \frac{- 1 + 2 + 8}{2} = \frac{9}{2}$

Vertex $\left(\frac{1}{2} , \frac{9}{2}\right)$

Y- intercept
At $x = 0$

$y = - 2 {\left(0\right)}^{2} + 2 \left(0\right) + 4$

$y = 4$

Y- intercept $\left(0 , 4\right)$

X- intercept

At $y = 0$

$- 2 {x}^{2} + 2 x + 4 = 0$

$- 2 {x}^{2} + 4 x - 2 x + 4 = 0$

$- 2 x \left(x - 2\right) - 2 \left(x - 2\right) = 0$

$\left(- 2 x - 2\right) \left(x - 2\right) = 0$

$- 2 x - 2 = 0$
$x = \frac{2}{- 2} = - 1$

X - intercept $\left(- 1 , 0\right)$

$x - 2 = 0$

$x = 2$

X - intercept $\left(2 , 0\right)$