This note covers rational functions of the form \(\dfrac{ax+b}{cx+d}\) and \(\dfrac{ax^2+bx+c}{dx+e}\), including their vertical, horizontal and oblique asymptotes, intercepts and graphs. These curves model quantities that level off or blow up, such as concentration, cost per unit and resistance.
Key ideas
A rational function is a polynomial divided by a polynomial. Two features dominate its graph: the places where the denominator is zero, and what happens when \(x\) becomes very large.
Linear over linear. Take \(f(x)=\dfrac{ax+b}{cx+d}\) with \(c\ne 0\) and \(ad-bc\ne 0\). It has a vertical asymptote where the denominator is zero, at \(x=-\dfrac dc\). As \(x\to\pm\infty\), the constants \(b\) and \(d\) matter less and less, so \(f(x)\to\dfrac ac\). This gives the horizontal asymptote \(y=\dfrac ac\). The domain excludes \(x=-\dfrac dc\), and the range excludes \(y=\dfrac ac\). The graph has two separate branches. Dividing the numerator by the denominator rewrites it as \(f(x)=\dfrac ac+\dfrac{k}{x+d/c}\), which is the graph of \(y=\dfrac1x\) stretched, reflected if \(k\lt 0\), and translated.
Quadratic over linear. Take \(g(x)=\dfrac{ax^2+bx+c}{dx+e}\). The vertical asymptote is again where the denominator is zero, at \(x=-\dfrac ed\), provided the numerator is not zero there. Since the numerator has a higher degree than the denominator, there is no horizontal asymptote. Instead, divide to write \(g(x)=mx+n+\dfrac{r}{dx+e}\). As \(x\to\pm\infty\) the fraction tends to zero, so the graph approaches the straight line \(y=mx+n\). This line is the oblique (slant) asymptote.
Intercepts. For the \(y\)-intercept, put \(x=0\). For the \(x\)-intercepts, set the numerator equal to zero (provided the denominator is not also zero at that value).
Sketching. Draw the asymptotes as dashed lines and mark the intercepts. For each branch, decide which side of the asymptote the curve lies on by testing a value of \(x\) close to the vertical asymptote, or by checking the sign of the expression. Your GDC can plot the graph, find intercepts, and locate maximum and minimum points. A graph can cross a horizontal or oblique asymptote, but it can never cross a vertical asymptote.
A rational function has no vertical asymptote at an \(x\)-value where the numerator and denominator share a factor. For example, \(\dfrac{x^2-1}{x-1}=x+1\) for \(x\ne 1\), so the graph is a straight line with a single missing point (a hole) at \(x=1\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Vertical asymptote | A line \(x=p\) that the graph approaches as \(y\to\pm\infty\). It occurs where the denominator is zero but the numerator is not. |
| Horizontal asymptote \(y=\dfrac ac\) for \(\dfrac{ax+b}{cx+d}\) | The value \(f(x)\) approaches as \(x\to\pm\infty\). Use the ratio of the \(x\) coefficients. |
| Oblique asymptote \(y=mx+n\) | For quadratic over linear: the quotient obtained by dividing the numerator by the denominator. |
| \(x=-\dfrac dc\) for \(\dfrac{ax+b}{cx+d}\) | Position of the vertical asymptote of a linear over linear function. |
| \(f(x)=\dfrac ac+\dfrac{k}{x+d/c}\) | Division form. Shows the transformation of \(y=\dfrac1x\). |
| \(x\)-intercepts: numerator \(=0\); \(y\)-intercept: \(f(0)\) | Used to place the graph on the axes. |
Worked examples
Example 1 Let \(f(x)=\dfrac{3x-1}{x+2}\). Find the equations of the asymptotes, the axis intercepts, and the domain and range.
The denominator is zero when \(x=-2\), and the numerator is \(-7\) there, so \(x=-2\) is a vertical asymptote.
For large \(|x|\), the function is close to \(\dfrac{3x}{x}=3\). So \(y=3\) is the horizontal asymptote.
To confirm, divide: \(3x-1=3(x+2)-7\), so \(f(x)=3-\dfrac{7}{x+2}\). The fraction tends to 0, which leaves 3.
For the \(y\)-intercept, \(f(0)=\dfrac{-1}{2}=-\tfrac12\). For the \(x\)-intercept, \(3x-1=0\) gives \(x=\tfrac13\).
The domain is all real \(x\) except \(-2\). From the division form, \(\dfrac{7}{x+2}\) can take any non-zero value, so \(f(x)\) takes any value except 3.
Answer: VA \(x=-2\); HA \(y=3\); intercepts \(\left(0,-\tfrac12\right)\) and \(\left(\tfrac13,0\right)\); domain \(x\ne-2\); range \(y\ne 3\)
Example 2 Let \(g(x)=\dfrac{x^2+x-6}{x-1}\). Find all the asymptotes and intercepts.
The denominator is zero at \(x=1\). The numerator there is \(1+1-6=-4\), which is not zero, so \(x=1\) is a vertical asymptote.
Divide the numerator by \(x-1\). Note that \((x-1)(x+2)=x^2+x-2\). The numerator is 4 less than this, so \(x^2+x-6=(x-1)(x+2)-4\).
Therefore \(g(x)=x+2-\dfrac{4}{x-1}\).
As \(x\to\pm\infty\), \(\dfrac{4}{x-1}\to 0\), so the graph approaches the line \(y=x+2\). This is the oblique asymptote. There is no horizontal asymptote.
For the \(y\)-intercept, \(g(0)=\dfrac{-6}{-1}=6\).
For the \(x\)-intercepts, solve \(x^2+x-6=0\). Factorise: \((x+3)(x-2)=0\), so \(x=-3\) or \(x=2\).
Answer: VA \(x=1\); oblique asymptote \(y=x+2\); intercepts \((0,6)\), \((-3,0)\), \((2,0)\)
Example 3 Let \(h(x)=\dfrac{x^2+1}{x-1}\). Find the asymptotes, then use your GDC to find the turning points and the range of \(h\).
The denominator is zero at \(x=1\) and the numerator is 2 there, so the vertical asymptote is \(x=1\).
Divide: \((x-1)(x+1)=x^2-1\), so \(x^2+1=(x-1)(x+1)+2\). Hence \(h(x)=x+1+\dfrac{2}{x-1}\), and the oblique asymptote is \(y=x+1\).
Plot \(y=h(x)\) on your GDC. The right-hand branch has a minimum point and the left-hand branch has a maximum point. Use the minimum and maximum features of your GDC.
The GDC gives the minimum at \((2.41,\,4.83)\) and the maximum at \((-0.414,\,-0.828)\).
Check exactly: the turning points occur at \(x=1\pm\sqrt2\), where \(h=2\pm2\sqrt2\). For \(x=1+\sqrt2\): \(h=2+\sqrt2+\dfrac{2}{\sqrt2}=2+2\sqrt2=4.83\).
Between the two turning-point values, \(-0.828\lt y\lt 4.83\), there are no points on the graph. The function takes values either at or below the maximum, or at or above the minimum.
Answer: VA \(x=1\); oblique asymptote \(y=x+1\); range \(y\le 2-2\sqrt2\) or \(y\ge 2+2\sqrt2\), that is \(y\le-0.828\) or \(y\ge 4.83\)
Common errors
- Writing the horizontal asymptote of \(\dfrac{3x-1}{x+2}\) as \(y=-\tfrac12\) or \(y=3x\). It is \(y=\dfrac{3}{1}=3\), the ratio of the coefficients of \(x\).
- Giving the vertical asymptote as \(x=2\) when the denominator is \(x+2\). Solve \(x+2=0\) to get \(x=-2\).
- Claiming a horizontal asymptote for quadratic over linear. There is an oblique asymptote instead, found by division.
- Saying the graph can never cross a horizontal or oblique asymptote. It can. Only vertical asymptotes are never crossed.
- Forgetting to give the asymptotes as equations (\(x=\ldots\) or \(y=\ldots\)). A number alone is not an equation of a line.
Exam tip: When a sketch is requested, label each asymptote with its equation and each intercept with its coordinates. Show where the curve approaches each asymptote from, rather than only drawing the shape.
Check yourself
- Find the vertical and horizontal asymptotes of \(f(x)=\dfrac{2x+5}{x-1}\).
Answer
The denominator is zero at \(x=1\), so the vertical asymptote is \(x=1\). The horizontal asymptote is \(y=\dfrac21=2\). - Find the intercepts of \(f(x)=\dfrac{2x+5}{x-1}\).
Answer
\(f(0)=\dfrac{5}{-1}=-5\), so the \(y\)-intercept is \((0,-5)\). Setting \(2x+5=0\) gives \(x=-2.5\), so the \(x\)-intercept is \((-2.5,0)\). - Find the oblique asymptote of \(g(x)=\dfrac{2x^2-3x+5}{x-2}\).
Answer
\((x-2)(2x+1)=2x^2-3x-2\), so \(2x^2-3x+5=(x-2)(2x+1)+7\). Then \(g(x)=2x+1+\dfrac{7}{x-2}\), and the oblique asymptote is \(y=2x+1\). - State the range of \(f(x)=\dfrac{3x-1}{x+2}\).
Answer
Writing \(f(x)=3-\dfrac{7}{x+2}\), the fraction can never be zero, so \(f(x)\) can never equal 3. The range is all real numbers except 3, that is \(y\ne3\).