If the function f(x) = ax^3 + bx^2 + 1 has a point of inflection at (-1,2).

At the point of inflection the second derivative is equal to 0.

f(x) = ax^3 + bx^2 + 1

=> f'(x) = 3ax^2 + 2bx

=> f''(x) = 6ax + 2b

6ax + 2b = 0

As the point of inflection is at (-1 , 2)

6a*(-1) + 2b = 0

=> -6a + 2b = 0

=> 3a = b

Also, f(-1) = 2

=> a(-1) + 3a(1) + 1 = 2

=> -a + 3a + 1 = 2

=> 2a = 1

=> a = 1/2

b = 3a = 3/2

**The value of a = 1/2 and b = 3/2**

Since the function is a polynomial, then it is continuous and differentiable.

Since (-1,2) is an inflection point, then x = -1 is the root of the 2nd derivative. We'll calculate the first derivative:

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

Now, we'll calculate the second derivative:

f"(x) = 6ax + 2b

f"(-1) = 0

-6a+2b=0

-6a=-2b

a = b/3

Since the point of inflection is on the graph, then (-1,2) verifies the function:

f(-1) = 2

-a+b+1 = 2

-a+b=1

-b/3 +b = 1

-b + 3b = 3

2b = 3

b = 3/2

a = 1/2

**The values of a and b are: a = 1/2 and b = 3/2.**