Question

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

Answer 1

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 + 2x - 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 = `(-1)^2 + 2(-1) - 8`
= -9