# How do you find the vertex and the intercepts for f(x)=-2x^2+2x+8?

Oct 3, 2017

Vertex is at $\left(0.5 , 8.5\right)$, y intercept is at $\left(0 , 8\right)$ and
x intercepts are at
$\left(- 1.56 , 0\right) \mathmr{and} \left(2.56 , 0\right)$

#### Explanation:

$f \left(x\right) = - 2 {x}^{2} + 2 x + 8 \mathmr{and} f \left(x\right) = - 2 \left({x}^{2} - x\right) + 8$ or

$f \left(x\right) = - 2 \left\{{x}^{2} - x + {\left(\frac{1}{2}\right)}^{2}\right\} + \frac{1}{2} + 8$ or

$f \left(x\right) = - 2 {\left(x - \frac{1}{2}\right)}^{2} + \frac{1}{2} + 8$ or

$f \left(x\right) = - 2 {\left(x - 0.5\right)}^{2} + 8.5$ . Comparing with vertex form of

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

here $h = 0.5 , k = 8.5 \therefore$ Vertex is at $\left(0.5 , 8.5\right)$ . y intercept

is found by putting $x = 0$ in the equation $y = - 2 {x}^{2} + 2 x + 8$ or

$y = 8 \therefore$ y intercept is $y = 8 \mathmr{and} \left(0 , 8\right)$

x intercept is found by putting $y = 0$ in the equation

$y = - 2 {x}^{2} + 2 x + 8 \mathmr{and} - 2 {x}^{2} + 2 x + 8 = 0$ or

$- 2 {\left(x - 0.5\right)}^{2} + 8.5 = 0 \mathmr{and} 2 {\left(x - 0.5\right)}^{2} = 8.5$ or

${\left(x - 0.5\right)}^{2} = 4.25 \mathmr{and} \left(x - 0.5\right) = \pm \sqrt{4.25}$ or

$x = 0.5 \pm \sqrt{4.25} \mathmr{and} x \approx 2.56 \left(2 \mathrm{dp}\right) , x \approx - 1.56 \left(2 \mathrm{dp}\right)$

x intercepts are at $\left(- 1.56 , 0\right) \mathmr{and} \left(2.56 , 0\right)$

graph{-2x^2+2x+8 [-20, 20, -10, 10]} [Ans]