Question

How do you find the vertex and x intercept of `f(x)=x^2+11x+30`?

Answer 1

Vertex `(-11/2, -1/4)`
x-intercepts: -5 and -6

Explanation:

`f(x) = x^2 + 11x + 30.`
x-coordinate of vertex:
`x = -b/(2a) = - 11/2`
y-coordinate of vertex:
`y(-11/2) = 121/4 - 121/2 + 30 = -121/4 + 120/4 = - 1/4`
Vertex `(-11/2, -1/4)`
To find the 2 x-intercepts, solve the quadratic equation f(x) = 0.
Find 2 real roots, knowing sum (-b = -11) and product (c = 30).
The 2 x-intercepts are: - 5 and -6 -> sum (- 11 = -b) --> product (30).