What is the equation of a curve with gradient 4x^3 -7x + 3/2 which passes through the point (2,9)?

The important thing to note is that examiners always give the minimum amount of information for the question to still be answerable. If you have not used all the information then you probably haven't answered it right.

Firstly, we must remember that the gradient of a curve is just the derivative of it's original equation. So what we want to do is reverse this process and get back to the original equation. The opposite of differentiating is to integrate.

Lets call our original equation f(x).
f(x) = integral(gradient)dx
so
f(x) = integral(4x³ - 7x + 3/2)dx

Remembering the standard rule: integral(xⁿ)dx = 1/(n+1)*xⁿ⁺¹ 
Or in words, we increase the power of x by 1 and then divide by this new power.

f(x) = 4*(x³⁺¹)/(3+1) - 7(x¹⁺¹)/(1+1) + 3/2*(x⁰⁺¹)/(0+1) + c   Notice here that x⁰=1 (anything to the power of 0 =1)
f(x) = x⁴ -7x²/2 + (3/2)x + c

The easiest mistake to make here is to miss out the "+ c" on the end. When you integrate without limits to the integral there must always be a constant at the end. The terms with an x tell us about the shape of the line, the constant fixes it's height on the y axis. If the "+ c" was left off I would be automatically asuming that it intercepts the y-axis at (0,0) this might not be true.

Now for the last part, we use the coordinates provided in the question to work out a value for c.
We know that when x =2, f(x) = 9
So, 9 = 2⁴- (7/2)*2² + (3/2)*2 + c
9 = 5 + c
This can only be true if c = 4
So our final equation is f(x) = x⁴ - (7/2)x² + (3/2)x +4