Vertical and oblique asymptotes

Text

Find the vertical and oblique asymptotes of the function $f(x)=\dfrac{2x^{2}+1}{x-3}$.

Solution

Vertical asymptote. The denominator vanishes at $x=3$, where the numerator equals $19\ne 0$: so $x=3$ is a vertical asymptote (indeed $\lim_{x\to 3^\pm}f(x)=\pm\infty$). Oblique asymptote. Perform the division: $\frac{2x^{2}+1}{x-3}=2x+6+\frac{19}{x-3}.$ As $x\to\pm\infty$ the remainder $\tfrac{19}{x-3}\to 0$, so the line $y=2x+6$ is an oblique asymptote, with slope $m=2$ and intercept $q=6$. $\boxed{\,x=3\ \text{(vertical)},\quad y=2x+6\ \text{(oblique)}\,}$