If the graph of y=x^3+ax^2+bx-4 has a point of infection at (1,-6), what is the value of b?

Question

anonymous · Answer

we should substitute x=1 and y=-6 i think.

precal · Answer

but that won't give me anything else, I need to take a derivative.  Point of inflection is 2nd derivative

anonymous · Answer

oh yes,my bad ;)

precal · Answer

ok gotta go, be back soon, I have to take someone to work

ganeshie8 · Answer

inflection at x = k   $\implies $  f''(k) = 0

precal · Answer

is that the first derivative at 0? @ganeshie8

ganeshie8 · Answer

second derivative is 0 at the point of inflection

precal · Answer

ok I am given that point (1,-6) and the second derivative is 
y " = 6x+2a

precal · Answer

I know the answer is 0 but I don't know why

ganeshie8 · Answer

y'' at x = 1 is 0 :

6(1) + 2a = 0

precal · Answer

I solved for a
a=-3
I went back to the first derivative and tried to solve for b and got b=3
but key states it is 0

larseighner · Answer

If you have a, why plug into the derivative.  I find b from plugging into the original function.

precal · Answer

because I don't really have a clue how to do this at all but let me go use the original function

precal · Answer

ok thanks now I see that b=0  @ganeshie8 @LarsEighner

ganeshie8 · Answer

exactly ! we're given (1, f(1))
not(1, f'(1))

so plugging in f'(x) for solving b fails...