Guided Example #1 - Rational Functions

The denominator ${\displaystyle x+3}$ has a value of ${\displaystyle 0}$if ${\displaystyle x=-3}$. So the line ${\displaystyle x=-3}$ is a vertical asymptote.

To find any horizontal asymptotes, rewrite the function by dividing the numerator
and denominator by the highest power of x:
${\displaystyle f\left(x\right)=\frac{2x+1}{x+3}=\frac{x(2+\frac{1}{x})}{x(1+\frac{3}{x})}=\frac{(2+\frac{1}{x})}{(1+\frac{3}{x})}}$

As ${\displaystyle x\to \infty }$ or${\displaystyle x\to -\infty }$the value of ${\displaystyle \frac{1}{x}}$ and ${\displaystyle \frac{3}{x}}$ approach ${\displaystyle 0}$, so
${\displaystyle f\left(x\right)\to \frac{2+0}{1+0}=2}$.

The denominator ${\displaystyle x+3}$ has a value of ${\displaystyle 0}$if ${\displaystyle x=-3}$. So the line ${\displaystyle x=-3}$ is a vertical asymptote.

To find any horizontal asymptotes, rewrite the function by dividing the numerator
and denominator by the highest power of x:
${\displaystyle f\left(x\right)=\frac{2x+1}{x+3}=\frac{x(2+\frac{1}{x})}{x(1+\frac{3}{x})}=\frac{(2+\frac{1}{x})}{(1+\frac{3}{x})}}$

As ${\displaystyle x\to \infty }$ or${\displaystyle x\to -\infty }$the value of ${\displaystyle \frac{1}{x}}$ and ${\displaystyle \frac{3}{x}}$ approach ${\displaystyle 0}$, so
${\displaystyle f\left(x\right)\to \frac{2+0}{1+0}=2}$.