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

Aug 16, 2016

vertex is (-1;-9), y-intercept is (0;-8), x-intercepts are (2;0) and (-4;0)

#### Explanation:

y- intercept is when x = 0, i.e y = 8
for x-intercepts, y = 0 , so ${x}^{2} + 2 x - 8$ = 0
i.e (x - 2)(x + 4) = 0
thus x = 2 or x = - 4.
The vertex has x-co-ordinate halfway between the x-intercept, x = -1
if x = -1 , y = ${\left(- 1\right)}^{2} + 2 \left(- 1\right) - 8$
= -9