Question

Ive been trying to solve this problem for forever!!! Pls help

Answer 1

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:

$\lim_{x \to 2^-} \dfrac{x^2-4}{x-2}$ must equal $\lim_{x \to 2^-} \left(ax^2 - bx-16 \right)$

This must also equal $f(2) = a(2)^2 -b(2)-16$  or $f(2) = 4a-2b-16$.

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:

      $\lim_{x \to 3^-} \left(ax^2-bx-16\right) = \lim_{x \to 3^+} \left(10x -a+b \right) = f(3)$

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.