Answer
$a = \frac{1}{2}$
$b = \frac{1}{2}$
Work Step by Step
$f(x) = \frac{x^2-4}{x-2}~~~$ if $x \lt 2$
$f(x) = ax^2-bx+3~~~$ if $2 \leq x \lt 3$
$f(x) = 2x-a+b~~~$ if $x \geq 3$
For the function to be continuous at $x=2$:
$\lim\limits_{x \to 2^-}\frac{x^2-4}{x-2} = f(2)$
$\lim\limits_{x \to 2^-}\frac{(x-2)(x+2)}{x-2} = a(2)^2-b(2)+3$
$\lim\limits_{x \to 2^-}(x+2) = 4a-2b+3$
$4 = 4a-2b+3$
$4a-2b-1=0$
$-8a+4b+2=0$
For the function to be continuous at $x=3$:
$\lim\limits_{x \to 3^-}ax^2-bx+3 = f(3)$
$a(3)^2-b(3)+3 = 2(3)-a+b$
$9a-3b+3 = 6-a+b$
$10a-4b-3 = 0$
We can add these two equations:
$-8a+4b+2=0$
$10a-4b-3 = 0$
$2a-1 = 0$
$a = \frac{1}{2}$
We can find $b$:
$10a-4b-3 = 0$
$10(\frac{1}{2})-4b-3 = 0$
$5-4b-3 = 0$
$-4b = -2$
$b = \frac{1}{2}$
