# How do you find the vertex and intercepts for y=2x^2+3x-8?

May 30, 2018

Vertex $\left(- \frac{3}{4} , - \frac{73}{8}\right)$
Y-Intercept $\left(0 , - 8\right)$
x-intercept ((sqrt73-3)/4,0); ((-sqrt73-3)/4,0)

#### Explanation:

Given -

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

$x = \frac{- b}{2 a} = \frac{- 3}{2 \times 2} = \frac{- 3}{4} = - \frac{3}{4}$

At x=-3/4; y=2(-3/4)^2+3(-3/4)-8

$y = 2 \left(- \frac{9}{16}\right) - \frac{9}{4} - 8$

$y = \frac{18}{16} - \frac{9}{4} - 8 = \frac{18 - 36 - 128}{16} = - \frac{146}{16} = - \frac{73}{8}$

Vertex

$\left(- \frac{3}{4} , - \frac{73}{8}\right)$

Intercept

Y-Intercept

At x=0; y=2(0^2)+3(0)-8=-8

$\left(0 , - 8\right)$

X-intercept

At y=0;0=2x^2+3x-8

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

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

${x}^{2} + \frac{3}{2} x = 4$

${x}^{2} + \frac{3}{2} x + \frac{9}{16} = 4 + \frac{9}{16} = \frac{64 + 9}{16} = \frac{73}{16}$

${\left(x + \frac{3}{4}\right)}^{2} = \pm \sqrt{\frac{73}{16}}$

$x + \frac{3}{4} = \pm \sqrt{\frac{73}{16}} = \pm \frac{\sqrt{73}}{4}$

$x = \frac{\sqrt{73}}{4} - \frac{3}{4} = \frac{\sqrt{73} - 3}{4}$

$x = - \frac{\sqrt{73}}{4} - \frac{3}{4} = \frac{- \sqrt{73} - 3}{4}$

x-intercept ((sqrt73-3)/4,0); ((-sqrt73-3)/4,0)