Answer
See explanation
Work Step by Step
If $Q(x_0)=0$, then the denominator of $y$ becomes zero at $x=x_0$, which may cause a vertical asymptote (the function “blows up”) or a hole (removable discontinuity), depending on whether $p(x_0)$ is also zero.
We have two main cases:
Case 1: $p(x_0)\not=0$
Denominator $= 0$, numerator $\not=0$
The function tends to $\pm\infty$ as $x\rightarrow x_0$
Graph behavior: vertical asymptote at $x=x_0$
Example:
$$y=\frac{1}{x-3}$$
Here $q(x)=x-3, q(3)=0, p(3)=1\not=0$.
As $x\rightarrow 3^-,y\rightarrow -\infty$; as $x\rightarrow 3^+,y\rightarrow +\infty$.
The graph has a vertical asymptote $x=3$.
Case 2: $p(x_0)=0$
Both numerator and denominator $= 0 \rightarrow$ indeterminate form $0/0$.
The behavior depends on factorization and multiplicity of the zero:
a) Common factor (removable discontinuity):
- If $p(x)=(x-x_0)^kr(x),q(x)=(x-x_0)^ks(x)$, we simplify by $x=x_0$.
- The graph has a hole at $x=x_0$, but no vertical asymptote.
- Example: $y=\frac{(x-1)(x+2)}{(x-1)(x-7)}$.
b) Different multiplicities:
- If the zero of $p(x)$ has lower multiplicity than zero of $q(x)$, vertical asymptote occurs.
- If the zero of $p(x)$ has higher multiplicity, function goes to $0$ at $x=x_0$.
