Question

How do you find the vertex and the intercepts for `f(x) = x^2 + 7x -8`?

Answer 1

Vertex `(-7/2, -81/4)`

Explanation:

`f(x) = x^2 + 7x - 8`
x-coordinate of vertex:
`x = -b/(2a) = -7/2`
y-coordinate of vertex:
`y(-7/2) = (49/4) + 7(-7/2) - 8 = 49/4 - 49/2 - 8 = -81/4`
Vertex (-7/2, - 81/4)
Make x = 0 --> y-intercept = - 8
To find x-intercepts, make x = 0 and solve f(x) = 0
Since a + b + c = 0, the, the 2 x-intercepts (real roots) are:
1 and `c/a = -8`
graph{x^2 + 7x - 8 [-40, 40, -20, 20]}