Calculus I: with the function f(x)=ax^3+bx^2+cx+d, find the values of a, b, c and d such that....

With the function f(x)=ax^3+bx^2+cx+d, find the values of a, b, c and d such that....
a.) f(0)=0
b.) x= 0, 2 are critical points
c.) x=1 is an inflection point
d.) f '(3)= 3/2
So I know I have to find the derivative and second derivative of f(x), so...
f '(x)= 3ax^2+2bx+c
and f ''(x)= 6ax+2b
I tried to solve for b in f ''(x) and then plugging 1 into x to get b=3a and then using THAT to solve f '(x).... but it didn't work. Am I at least on the right track? And if so, can someone please help me get further? I'm just so stuck and frustrated with this problem.
Thank you!

f(x) = ax^3 + bx^2 + cx + d

f(0) = a(0^3) + b(0^2) + c(0) + d = 0
0 + 0 + 0 + d = 0
d = 0....(1)

f'(x) = 3ax^2 + 2bx + c
f''(x) = 6ax + 2b

from c.) x = 1 is an inflection point, so
f''(1) = 0
f''(1) = 6a(1) + 2b = 0
6a + 2b = 0
a = -b/3 ----> same idea, you were on the right track, except you forgot the negative sign.

from b.) x = 0, 2 are critical points
f'(x) = 3ax^2 + 2bx + c = 0
f'(0) = 0 + 0 + c = 0
c = 0
f'(2) = 3a(4) + 2b(2) + c = 0
f'(2) = 12a + 4b + 0 = 0
12a = -4b
a = -b/3

and you also know that f'(3) = 3/2 from d.)
f'(3) = 27a + 6b = 3/2
54a + 12b = 3
18a + 4b = 1

You have two equations:
18a + 4b = 1
a = -b/3

Two equations, two unknowns, solve for a, b.

18(-b/3) + 4b = 1
-6b + 4b = 1
-2b = 1
b = -1/2

a = (-1/2) / (-3)
a = 1/6

Thus,

a = 1/6, b = -1/2, c = 0, d = 0

And so

f(x) = (1/6)x^3 - (1/2)x^2

Hope this helps :D

A) f(0)=0, so d=0

B) f '(0) = 0 and f '(2)=0
c=0 and 0= 3a(4)+2b(2)
12a+4b=0
3a+b=0
b=-3a

C) f " ( 1)=0
6a+2b=0
b=-3a

D) 3a(9)+2b(3)=3/2

27a +6b=3/2
27a+ 6(-3a)=3/2
9a= 3/2
a= 1/6
b=-1/2

So y = (1/6)x^3 -(1/2)x^2

Hoping this helps!

f(0) = 0 ---> d = 0
f '(0) = 0 = f '(2)
f "(1)=0
f '(3)=3/2