Solving for the constants of a cubic function where its graph is tangent to the x-axis at x=3

Question

$$x^3-4x^2+ax+b$$

I am unsure of my understanding of theory yet, but from what I do understand I assume that y=0 when x=3 because the graph is tangent to the x-axis at that point. With that amount of information what can I do to find the real constants a and b?

Answer 1

Solution without derivatives.

We need that $3$ would be a root of degree two.

Thus, $$3^3-4\cdot3^2+3a+b=0,$$ which gives $$b=9-3a$$ and we obtain: $$x^3-4x^2+ax+9-3a=0$$ or $$x^3-4x^2+9+a(x-3)=0$$ or $$x^3-3x^2-x^2+3x-3x+9+a(x-3)=0$$ or $$(x-3)(x^2-x-3+a)=0,$$ which gives $$3^2-3-3+a=0,$$ $$a=-3$$ and $$b=18.$$

Answer 2

If $p(x)=x^3-4x^2+ax+b$, you want to have $p(3)=0$ and $p'(3)=0$. But$$p(3)=0\iff27-36+3a+b=0$$and$$p'(3)=0\iff27-24+a=0.$$Can you take it from here?