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

1 Answer
Jun 20, 2016

Vertex is at $\left(- 1 , - 9\right)$ y intercept is at $\left(0 , - 8\right)$ x intercepts are at (-4,0) & (2,0)

Explanation:

$F \left(x\right) = {x}^{2} + 2 x - 8 = {x}^{2} + 2 x + 1 - 9 = {\left(x + 1\right)}^{2} - 9 \therefore$Vertex is at $\left(- 1 , - 9\right)$
Putting x= 0 we get y-intercept as$F \left(x\right) = - 8$ and putting $F \left(x\right) = 0$
we get x -intercepts as ${x}^{2} + 2 x - 8 = 0 \mathmr{and} \left(x + 4\right) \left(x - 2\right) = 0 \mathmr{and} x = - 4 \mathmr{and} x = 2$ graph{x^2+2x-8 [-40, 40, -20, 20]} [Ans]