# How do you find the intercepts, vertex and graph f(x)=-2x^2+8x-3?

Feb 21, 2017

Vertex $\left(2 , 5\right)$
X intercept (3.58, 0); (0.42, 0)
Y-Intercept $\left(0 , - 3\right)$

#### Explanation:

Given -

$f \left(x\right) = - 2 {x}^{2} + 8 x - 3$

Vertex

$x = \frac{- b}{2 a} = \frac{- 8}{2 \times - 2} = \frac{8}{- 4} = - 2$

At $x = 2$

$y = - 2 \left({2}^{2}\right) + 8 \left(2\right) - 3$
$y = - 8 + 16 - 3 = 16 - 11 = 5$

Vertex $\left(2 , 5\right)$

X intercept

$- 2 {x}^{2} + 8 x - 3 = 0$

${x}^{2} - 4 x + \frac{3}{2} = 0$ [divide all the terms by $- 2$]

${x}^{2} - 4 x = - \frac{3}{2}$ [Take the constant term to the right]
${x}^{2} - 4 x + 4 = - \frac{3}{2} + 4$ (divide the coefficient of $x$ square it and add to both sides]
${x}^{2} - 4 x + 4 = - \frac{3}{2} + 4 = \frac{- 3 + 8}{2} = \frac{5}{2}$
${\left(x - 2\right)}^{2} = - \frac{5}{2}$
$x - 2 = \pm \sqrt{\frac{5}{2}} = \pm 1.58$

$x = 1.58 + 2 = 3.58$
$x = - 1.58 + 2 = 0.42$

X intercept (3.58, 0); (0.42, 0)

Y-Intercept
At $x = 0$

$y = - 2 {\left(0\right)}^{2} + 8 \left(0\right) - 3$
$y = - 3$

Y-Intercept $\left(0 , - 3\right)$

To graph the function take a few values on either side of $x = 2$
Find the corresponding $y$ values.
Tabulate them. Graph the pairs