Oblique asymptotes of rational functionsAQA A-Level Further Maths: Revision notes

A rational function $y=\frac{p(x)}{q(x)}$ has an oblique (slanting) asymptote when the degree of the numerator is exactly one more than the degree of the denominator, for example $\frac{x^2+x-6}{x-1}$, a quadratic over a linear. Compare the degrees:

• numerator degree less than denominator: horizontal asymptote $y=0$;
• degrees equal, such as $\frac{ax^2+bx+c}{dx^2+ex+f}$: horizontal asymptote $y=\frac ad$;
• numerator degree one more: oblique asymptote $y=mx+c$;
• two or more higher: no straight-line asymptote. Vertical asymptotes still occur where the denominator is zero and the numerator is not.

Divide the numerator by the denominator. The quotient is $mx+c$ and the remainder is a constant $r$: $\frac{p(x)}{q(x)}=mx+c+\frac{r}{q(x)}.$ As $|x|\to\infty$ the fraction $\frac{r}{q(x)}\to0$, so $y=mx+c$ is the oblique asymptote. Example: $x^2+x-6=(x-1)(x+2)-4$, so $\frac{x^2+x-6}{x-1}=x+2-\frac{4}{x-1}$, with asymptotes $x=1$ and $y=x+2$. You can divide by long division, or by writing the numerator as $(x-1)(\ldots)+r$ and matching coefficients. A check: put in $x=2$: $\frac{4+2-6}{1}=0$ and $2+2-\frac41=0$.

The difference between the curve and its asymptote is the remainder term: $y-(mx+c)=\frac{r}{q(x)}$. Its sign tells you where the curve lies.

• If $\frac{r}{q(x)}>0$ the curve is above the asymptote; if $<0$ it is below.
• Because $\frac rq$ is never zero (when $r\ne0$), the curve never meets an oblique asymptote. Example: $y=2x-7+\frac{15}{x+2}$ is above $y=2x-7$ when $x>-2$ and below it when $x<-2$. To prove no intersection, set $2x^2-3x+1=(2x-7)(x+2)$ and obtain the contradiction $1=-14$.

To describe a curve fully, find: vertical asymptotes (denominator zero), the oblique or horizontal asymptote, the $y$-intercept (put $x=0$) and the $x$-intercepts (numerator zero). Example: $y=\frac{x^2-x-6}{x+1}=x-2-\frac{4}{x+1}$ has asymptotes $x=-1$ and $y=x-2$, meets the $y$-axis at $(0,-6)$ and the $x$-axis at $(3,0)$ and $(-2,0)$. A line parallel to the oblique asymptote meets the curve at most once, because the $x^2$ terms cancel when you equate: $\frac{2x^2+3}{x-1}=2x$ gives $3=-2x$, one point $\left(-\frac32,-3\right)$.

If the asymptote or a point on the curve is given, use division with letters. For $y=\frac{x^2+ax+b}{x-1}$: $x^2+ax+b=(x-1)(x+a+1)+(a+b+1),$ so $y=x+(a+1)+\frac{a+b+1}{x-1}$. An asymptote $y=x+3$ gives $a=2$. A point $(0,-2)$ gives $\frac{b}{-1}=-2$, so $b=2$. The remainder is $a+b+1=5$, so $y=x+3+\frac{5}{x-1}$ and the curve is above the asymptote for $x>1$.