1. Factor the numerator and denominator. 2. Identify zeros of the denominator. 3. Check if those zeros are also zeros of the numerator. 4. If a zero is only in the denominator, there's a vertical asymptote there.

1. Find the vertical asymptote. 2. Evaluate the one-sided limits as $x$ approaches the asymptote from the left and right. 3. Determine if the function approaches $\infty$ or $-\infty$ from each side.

1. Factor the numerator and denominator. 2. If a factor cancels out from both, there is a hole at the x-value that makes that factor zero.

1. Factor: $r(x) = \frac{(x-1)(x+1)}{(x-1)(x-1)}$. 2. Simplify: $r(x) = \frac{x+1}{x-1}$. 3. Vertical asymptote: $x=1$.

Vertical Asymptotes: Occur when the denominator is zero, and the numerator is non-zero. Function approaches infinity. | Holes: Occur when both numerator and denominator are zero. Represent removable discontinuities.

If $x = a$ makes the denominator $q(x)$ zero and the numerator $p(x)$ non-zero, then there's a potential vertical asymptote at $x = a$.

It might indicate a 'hole' in the graph rather than a vertical asymptote.

If the multiplicity of a zero in the denominator is greater than in the numerator, there's a vertical asymptote. Higher difference means faster approach to infinity.

As $x$ approaches a vertical asymptote from the left or right, the function approaches either positive or negative infinity.

One-sided limits approaching a vertical asymptote will tend towards $\infty$ or $-\infty$.