Rational Functions Math Example 4

Solution

1. 1
   Factor the numerator: $x^2 + 2x - 3 = (x + 3)(x - 1)$. So $f(x) = \frac{(x+3)(x-1)}{x-1}$.
2. 2
   The $(x-1)$ cancels: $f(x) = x + 3$ for $x \neq 1$. Since the factor cancels, $x = 1$ is a hole (not a vertical asymptote). Hole at $(1, 4)$.
3. 3
   No vertical asymptotes remain. The simplified form $x + 3$ is linear, so there is an oblique asymptote $y = x + 3$ (which the function equals everywhere except the hole). No horizontal asymptote.