2.13 Rational functions

A rational function is one polynomial divided by another. The simplest examples, $y = \frac{1}{x}$ and $y = \frac{ax + b}{cx + d}$

Rational functions model quantities that are ratios, such as the average cost of making an item, the concentration of a substance in a mixture, or the efficiency of a process. Their asymptotes describe the long-run behaviour of these models.

Rational functions and their domains

The two families studied here are

In the first the denominator has the higher degree; in the second the numerator does. This difference decides the long-run behaviour of the graph.

To sketch any rational function, find these features in turn:

1. The domain and the vertical asymptotes, from the zeros of the denominator.

2. The behaviour on each side of each vertical asymptote.

3. The horizontal or oblique asymptote, from the degrees of the numerator and the denominator.

4. The intercepts with the axes.

5. Any turning points, from a GDC or by differentiation.

Vertical asymptotes

A vertical asymptote occurs where the denominator is zero and the numerator is not. Near such a value the denominator is very small, so the fraction is very large in size: $f(x) \to \infty$ or $f(x) \to -\infty$.

For $f(x) = \frac{ax + b}{cx^2 + dx + e}$ the number of vertical asymptotes depends on the discriminant of the denominator, $\Delta = d^2 - 4ce$

A common factor gives a gap, not an asymptote

If the numerator and the denominator are both zero at $x = k$, they share the factor $(x - k)$. After cancelling it, the graph is that of the simplified function with the single point at $x = k$ missing. For example,

The graph has a vertical asymptote at $x = -2$ only. At $x = 2$ there is a gap in the curve at the point $\left(2, \frac{1}{4}\right)$

Behaviour on each side of a vertical asymptote

On each side of a vertical asymptote the graph goes either up to $+\infty$ or down to $-\infty$. The direction comes from the sign of $f(x)$ just to the left and just to the right of the asymptote, which in turn comes from the signs of the factors.

• At a simple zero of the denominator, such as $(x - 3)$, the denominator changes sign, so the graph goes to $+\infty$ on one side and $-\infty$ on the other.

• At a repeated zero, such as $(x - 1)^2$

To decide which way, substitute a value just beyond the asymptote, or reason with signs. The notation $x \to 3^+$ means "$x$ approaches $3$ from values greater than $3$", and $x \to 3^-$

Horizontal asymptotes

The long-run behaviour of a rational function, as $x \to \pm\infty$, depends on the degrees of the numerator and the denominator. For large $|x|$ only the leading terms matter, just as for polynomials.

For $f(x) = \frac{ax + b}{cx^2 + dx + e}$, divide the numerator and the denominator by $x^2$

So every function of this form has the horizontal asymptote $y = 0$.

Crossing a horizontal asymptote

An asymptote describes what happens as $x \to \pm\infty$, not what happens in the middle of the graph. A graph can cross its horizontal asymptote for moderate values of $x$, and for $f(x) = \frac{ax + b}{cx^2 + dx + e}$

To find which side of $y = 0$ the graph approaches from, look at the sign of $f(x)$ for large positive and large negative $x$. For $f(x) = \frac{2x - 1}{x^2 - x - 6}$

The sign of the function between key values

The zeros of the numerator and of the denominator split the $x$-axis into intervals. On each interval no factor changes sign, so $f(x)$ keeps one sign throughout, and a table of signs shows at once whether the graph is above or below the $x$-axis there. For $f(x) = \frac{2x - 1}{(x - 3)(x + 2)}$

The same table is the basis of the sign diagrams used to solve rational inequalities in 2.15.

Functions with no vertical asymptote

When the denominator has no real zeros, the graph is a single unbroken curve. It still has the horizontal asymptote $y = 0$, so it rises from the asymptote and returns to it: such graphs always have a maximum and a minimum, or a single turning point.

Finding the range without calculus

For a rational function with a quadratic part, the range can also be found algebraically. A value $k$ is in the range exactly when the equation $f(x) = k$ has a real solution. Clearing the fraction turns this into a quadratic in $x$, which has real solutions when its discriminant is at least zero.

For $f(x) = \frac{4x}{x^2 + 1}$, the equation $\frac{4x}{x^2 + 1} = k$

When $k \neq 0$ this quadratic has real roots if and only if $16 - 4k^2 \ge 0$, that is $-2 \le k \le 2$

A repeated factor in the denominator

Oblique asymptotes

When the numerator has degree exactly one more than the denominator, as in $f(x) = \frac{ax^2 + bx + c}{dx + e}$, the function grows without limit as $x \to \pm\infty$

Finding an oblique asymptote by division

Divide the numerator by the denominator (polynomial division, 2.12). Since the divisor is linear, the remainder $r$ is a constant, and

where $mx + k$ is the quotient. As $x \to \pm\infty$, the term $\frac{r}{dx + e} \to 0$

The same form shows on which side of the asymptote the graph lies. The vertical distance from the line to the curve is $\frac{r}{dx + e}$: where this is positive the curve is above the asymptote, and where it is negative the curve is below it.

Because $\frac{4}{x + 1} > 0$ for $x > -1$ and $< 0$

Turning points and range

Graphs with an oblique asymptote and one vertical asymptote have two branches. Each branch either has one turning point or none. The turning points can be found with a GDC, or by differentiating the divided form, which is much easier than using the quotient rule on the original fraction.

The discriminant method gives the same range without calculus. Setting $\frac{x^2 + 3x + 6}{x + 1} = k$ and clearing the fraction gives $x^2 + (3 - k)x + (6 - k) = 0$

that is, when $k \le -3$ or $k \ge 5$. Solving a quadratic inequality like this one is covered in 2.7 and 2.15.

When the turning points are not simple numbers

Many rational functions have turning points with surd coordinates. Calculus gives exact values; a GDC gives decimal approximations, which are acceptable when a calculator is allowed and exact values are not asked for.

Finding a rational function from its features

The divided form $f(x) = mx + k + \frac{r}{x - h}$ makes it easy to build a function with given asymptotes: the vertical asymptote is $x = h$

A parameter in the denominator can change the number of vertical asymptotes. For $f(x) = \frac{x + 1}{x^2 + kx + 4}$, the denominator has discriminant $k^2 - 16$

• If $-4 < k < 4$, the discriminant is negative: no vertical asymptotes.

• If $k = \pm 4$, the denominator is $(x \pm 2)^2$

Rational functions as models

In a model, only the part of the graph with a sensible domain is used, and the asymptotes describe long-run behaviour in the context.

Average cost models give oblique asymptotes. If making $x$ items costs $0.02x^2 + 5x + 2000$ dollars (a fixed cost of 2000 dollars plus costs that rise faster than the number of items), the average cost per item is

For small production runs the term $\frac{2000}{x}$ dominates: the fixed cost is shared between few items. For large runs $A(x)$ approaches the oblique asymptote $A = 0.02x + 5$

Exam-style questions on rational functions

The following question is typical of Paper 1 (no GDC), where every feature must be found algebraically.

Self-check

• To do:

  I can find the vertical asymptotes of a rational function, after checking for common factors.

• To do:

  I can decide whether the graph goes to $+\infty$ or $-\infty$ on each side of a vertical asymptote.

• To do:

  I can find the horizontal asymptote by comparing degrees, and explain why the graph can cross it.

• To do:

  I can find an oblique asymptote by division and say on which side of it each branch lies.

• To do:

  I can find turning points with a GDC or by differentiating the divided form.

• To do:

  I can find the range of a rational function, including by the discriminant method.

• To do:

  I can sketch rational functions with all asymptotes, intercepts and turning points labelled.

• To do:

  I can find unknown coefficients from given asymptotes and points.