# How do you find the values of 'a' and 'b' that make f(x) continuous everywhere given (x^2 - 4)/(x - 2) if x < 2, ax^2 - bx + 3 if 2 < x < 3 and 2x - a + b if x>=3?

$f \left(x\right) = \left\{\begin{matrix}\frac{{x}^{2} - 4}{x - 2} \text{ & " & x < 2 \\ ax^2 - bx + 3" & " & 2 < x < 3 \\ 2x - a + b" & } & x \ge 3\end{matrix}\right.$
The way the function is defined, we already know that $x \in \left(- \infty , 2\right) \cup \left(2 , 3\right) \cup \left[3 , \infty\right)$. That is, $x \ne 2$ (but $x$ can be $3$). However, because of that, there is no way for $f \left(x\right)$ to be continuous everywhere.
$f \left(x\right)$ would have a discontinuity (a hole) at $x = 2$, since even though you can cancel out $x - 2$, you still have $2 - 2 = 0$.
$\therefore$ One might say that $f \left(x\right)$ is continuous when $a , b$ are real and finite, but only if $x \ne 2$. So strictly speaking, $f \left(x\right)$ cannot be continuous everywhere.