For the function to be continuous at any x-value you need the left-hand limit to match the right-hand limit to match the function's value at that x-value.
For example, for the function to be continuous at x=2:
[tex]\lim_{x \to 2^-} \dfrac{x^2-4}{x-2}[/tex] must equal [tex]\lim_{x \to 2^-} \left(ax^2 - bx-16 \right)[/tex]
This must also equal [tex]f(2) = a(2)^2 -b(2)-16[/tex] or [tex]f(2) = 4a-2b-16[/tex].
So start by finding the first limit that has no a's or b's in it and set that equal to 4a-2b-16.
The problem is that this is only one equation and there are two variables, so we need a second equation to set up to be able to solve for a and b.
So, you need to repeat that whole process with the pieces on either side of x=3. We need to have:
[tex]\lim_{x \to 3^-} \left(ax^2-bx-16\right) = \lim_{x \to 3^+} \left(10x -a+b \right) = f(3)[/tex]
That will give you a second equation with a's and b's. Once you have that, you'll have a system which you can solve using substitution or elimination.
