How do I find g(x) given the stationary points (2,9)?

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given the info I form the equation g(x) = AX³+BX²+AX+C and did g'(x) = 0 to find B = -13A/4 (when x = 2). So using B and putting it into g(x) = 9, I get AX³-13A/4X²+AX+C = 9 and I am confused on how to find the values for A, B and C...

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Following from your solving for B, we have the expression $$g(x)=Ax^3-\frac{13A}{4}x^2+Ax+C.$$ Since $g$ passes through the origin we have $$0=A\cdot0^3-\frac{13A}{4}\cdot0^2+A\cdot0+C.$$ giving $C = 0$. So far it reduces to $$g(x)=Ax^3-\frac{13A}{4}x^2+Ax.$$ Since $g$ passes through $(2,9)$ we have $$9=A\cdot2^3-\frac{13A}{4}\cdot2^2+A\cdot2= 8A-13A+2A = -3A.$$ Solving for $A$ and $B$ gives $A=-3$ thus $B=\frac{-3\cdot 13}{4}=-39/4.$ Finally $$g(x)=-3x^3-\frac{39}{4}x^2-3x.$$