Q.Find the asymptotes of the following curves:

Step 1 (i). $f(x)=\dfrac{x^2}{x^2-1}$.

Denominator zero at $x=\pm1$; numerator there is $1\ne0$, so $x=1,x=-1$ are vertical asymptotes.

Degrees equal (both 2): $\displaystyle\lim_{x\to\pm\infty}\frac{x^2}{x^2-1}=\lim_{x\to\pm\infty}\frac{1}{1-1/x^2}=1$. Horizontal asymptote $y=1$.

Step 2 (ii). $f(x)=\dfrac{x^2}{x+1}$.

Denominator zero at $x=-1$; numerator there is $1\ne0$, so $x=-1$ is a vertical asymptote.

Numerator degree (2) exceeds denominator (1) by exactly 1: slant asymptote. Divide: $x^2=(x+1)(x-1)+1$, so $f(x)=x-1+\dfrac{1}{x+1}$. Slant asymptote: $y=x-1$.

Step 3 (iii). $f(x)=\dfrac{3x}{\sqrt{x^2+2}}$.

Denominator $\sqrt{x^2+2}>0$ always (never zero), so there is no vertical asymptote.

As $x\to+\infty$: $\sqrt{x^2+2}=x\sqrt{1+2/x^2}$ (for $x>0$), so $f(x)=\dfrac{3x}{x\sqrt{1+2/x^2}}=\dfrac{3}{\sqrt{1+2/x^2}}\to3$.

As $x\to-\infty$: $\sqrt{x^2+2}=-x\sqrt{1+2/x^2}$ (for $x<0$, since $\sqrt{x^2}=|x|=-x$), so $f(x)=\dfrac{3x}{-x\sqrt{1+2/x^2}}=\dfrac{-3}{\sqrt{1+2/x^2}}\to-3$.

Horizontal asymptotes: $y=3$ (as $x\to\infty$) and $y=-3$ (as $x\to-\infty$).

Step 4 (iv). $f(x)=\dfrac{x^2-6x-1}{x+3}$.

Denominator zero at $x=-3$; numerator there is $9+18-1=26\ne0$, so $x=-3$ is a vertical asymptote.

Numerator degree exceeds denominator by 1: slant asymptote. Long division: $x^2-6x-1=(x+3)(x-9)+26$, so $f(x)=x-9+\dfrac{26}{x+3}$. Slant asymptote: $y=x-9$.

Step 5 (v). $f(x)=\dfrac{x^2+6x-4}{3x-6}$.

Denominator zero at $x=2$; numerator there is $4+12-4=12\ne0$, so $x=2$ is a vertical asymptote.

Numerator degree exceeds denominator by 1: slant asymptote. Dividing $x^2+6x-4$ by $3x-6$: quotient $\dfrac{x+8}{3}$ with remainder $12$, i.e. $f(x)=\dfrac{x+8}{3}+\dfrac{12}{3x-6}$. Slant asymptote: $y=\dfrac{x}{3}+\dfrac83$.

Vertical asymptotes from the denominator's zeros; horizontal/slant from the degree comparison or long division

• Forgetting the $|x|=-x$ subtlety for $x<0$ under a square root, which flips the sign of the $x\to-\infty$ horizontal asymptote
• Skipping the check that the numerator is nonzero at the denominator's zero before declaring a vertical asymptote