Solving for a and b in one equation

Suppose the function f(x) = 2x^3 + ax^2 + bx has a critical point at x =-1, for which f(-1) = -2. What are a, b?

I've tried solving this so many ways, but can't get the right answer. I know that for a critical point to exist, the derivative of the equation should equal 0 when x=-1. I did this to find what b equals with respect to a and then plugged it back in to find a, but it doesn't work. Thanks for any help.

f(x) = 2x^3+ ax^2+bx
f ' (x) = 6x^2+ 2ax+b
for x = -1 you have critical point then f '(-1) = 0 and f ( -1) = 2
then

f ' (-1) = 6(-1)^2+ 2a(-1)+b = 0 then 6 -2a+b = 0
f (-1) = 2(-1)^3+ a(-1)^2+b(-1) = 0 then -2+a-b = -2

Now solve
6 -2a+b = 0
-2+a-b = -2

- 2a+b = -6
a - b = 0
a = 6
b = 6
f(x) = 2x^3 + 6x^2 + 6x

We know two things:
f'(-1) = 0... from the definition of a critical point
f(-1) = -2... which is given

That's enough information to solve for a and b (two equations for two variables)

from the second, we have
f(-1) = 2(-1)^3 + a(-1)^2 + b(-1) = -2 --->
-2 + a - b = -2 -->
a = -b

from the first, we have
f'(1) = 6(-1)^2 + 2*a(-1) + b = 0 -->
6 - 2a + b = 0 -->
2a - b = 6... plugging in b = -a, we get:
2a + a = 6
3a = 6
a = 2 --> b = -2

and there's your answer.

OK, your approach sounds right. Let's see.
The derivative is 6x^2 + 2ax + b.
So we have simultaneous equations,
6 -2a + b = 0 and
2*(-1) + 1a -1b = -2.
From the first equation, b = 2a-6, so the second equation becomes
-2 + a - 2a + 6 = -2,
Then a =6 and b = 6, I think. Check it out.

f(-1)=-2 so 2(-1)^3+a(-1)^2+b(-1)=-2 simplify that a=b
take derivative f'(x)=6x^2+2ax+b=0 simplify b=2a-6 so a=2a-6 a=6 b=6

taking derivative
6*x^2+2*a*x+b=0
putting x=-1
6-2a+b=0
b-2a=(-6)..................eqn(1)

f(-1)=(-2)
-2+a-b=-2
a-b=0........................eqn(2)

solve eqn 1 and 2 you get answer.