I am to graph the function but i cant find where f''(x) equals zero

f(x)= x^4 - 4x^3 - 8x^2 + 48x

f'(x)= 4 (x-2)(x+2)(x-3)

f''(x)= 4(3x^2 -6x- 4)

my problem is i am unable to get the concavity.

what would be my intervals?

Re: I am to graph the function but i cant find where f''(x) equals zero

Have you tried factoring the 3x^2-6x-4 and then setting those to 0

Re: I am to graph the function but i cant find where f''(x) equals zero

Hello, Alioyo!

You're taking Calculus and you don't know how to solve a quadratic equation?

You have: .

Quadratic Formula: .

The intervals are: .