Question

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

Answer 1

Vertex is at `(-1, -9)` y intercept is at `(0,-8)` x intercepts are at `(-4,0) & (2,0)`

Explanation:

`F(x) = x^2+2x-8 = x^2+2x +1 - 9 =(x+1)^2 -9 :.`Vertex is at `(-1, -9)`
Putting x= 0 we get y-intercept as`F(x) = -8` and putting `F(x) =0`
we get x -intercepts as `x^2+2x-8=0 or (x+4)(x-2)=0 or x = -4 and x=2` graph{x^2+2x-8 [-40, 40, -20, 20]} [Ans]