Question

How do you find the vertex and the intercepts for `f(x) = -3x^2 - 6x - 2`?

Answer 1

Vertex (1, -11)
`x = -1 +- sqrt3/3`

Explanation:

x-coordinate of the vertex:
`x = -b/(2a) = 6/-6 = 1`
y-coordinate of vertex:
`y(1) = - 3 - 6 - 2 = -11`
Vertex (1, -11)
Make x = 0, y-intercept = - 2
To find x-intercepts, make y = 0 and solve the quadratic equation:
`f(x) = -3x^2 - 6x -2 = 0.`
`D = d^2 = b^2 - 4ac = 36 - 24 = 12` --> `d = +- 2sqrt3`
There are 2 x-intercepts (real roots):
`x = -b/(2a) +- d/(2a) = 6/-6 +- 2sqrt3/6 = -1 +- sqrt3/3`