Rational functions and asymptotes practice questions — IB Maths AA

$x^2 - 4 = 0 \Rightarrow x = \pm 2$. Both vertical asymptotes.

$f(x) = x + \dfrac{1}{x}$. As $x \to \pm\infty$, $\dfrac{1}{x} \to 0$, so the slant asymptote is $y = x$.

$\tan x = \dfrac{\sin x}{\cos x}$ has vertical asymptotes where $\cos x = 0$, i.e. $x = \dfrac{\pi}{2} + n\pi$.

$x - 1 \to 0^+$, so $\dfrac{1}{x - 1} \to +\infty$.

$\displaystyle\lim_{x \to \pm\infty} \dfrac{2x + 1}{x - 5} = 2$, so $y = 2$.

$\lim_{x \to \infty} x^2 = \infty$, so there is no horizontal asymptote.

$x - 3 = 0 \Rightarrow x = 3$.

$\dfrac{2x^2 + 3x + 1}{x - 1} = 2x + 5 + \dfrac{6}{x - 1}$. As $x \to \pm\infty$, the remainder vanishes, leaving the slant asymptote $y = 2x + 5$.

$\lim_{x \to \infty} e^{-x} = 0$, so $y = 0$ is a horizontal asymptote at $+\infty$.

Vertical: $x = \pm 1$ (two). Horizontal: $\lim f(x) = 1$ (one). Total: $3$. (No slant — the rational function has equal-degree numerator and denominator.)