2.8 Reciprocal and rational functions

Many quantities depend on each other so that when one increases, the other decreases in proportion. At a fixed temperature, squeezing a gas into half the volume doubles its pressure; driving a fixed distance twice as fast takes half the time. The function that describes these situations is the reciprocal function $f(x) = \frac{1}{x}$, and its graph has a shape quite different from the lines and parabolas met so far: two separate branches that approach, but never meet, a pair of lines called asymptotes.

The reciprocal function belongs to a larger family, the rational functions, which are quotients of polynomials. The rational functions $\frac{ax + b}{cx + d}$ studied here have graphs that are transformed copies of $y = \frac{1}{x}$, so the same key features (two asymptotes, two branches, no turning points) appear every time. Recognising those features quickly makes these functions easy to sketch, to invert and to use as models.

The reciprocal function

Definition and values

The reciprocal of a non-zero number $x$ is $\frac{1}{x}$: the number that multiplies by $x$ to give 1. The reciprocal of 4 is $\frac{1}{4}$, the reciprocal of $-\frac{2}{3}$ is $-\frac{3}{2}$, and the number 0 has no reciprocal, because no number multiplied by 0 gives 1.

A table of values shows how the reciprocal function behaves. Small inputs give large outputs, large inputs give small outputs, and every output has the same sign as its input.

Because $f(x)$ and $x$ always have the same sign, the graph lies only in the first quadrant (both positive) and the third quadrant (both negative). Because $f(0)$ is undefined, the graph does not cross the $y$-axis, and because $\frac{1}{x}$ is never 0, it does not cross the $x$-axis either.

Asymptotes of the reciprocal function

As $x$ takes positive values closer and closer to 0 (0.1, 0.01, 0.001, …), $\frac{1}{x}$ takes the values 10, 100, 1000, …, growing without limit. This is written $\frac{1}{x} \to \infty$ as $x \to 0^+$, where $0^+$ means "0 from the positive side". In the same way, $\frac{1}{x} \to -\infty$ as $x \to 0^-$. Near $x = 0$ the curve becomes almost vertical and runs alongside the $y$-axis without ever touching it.

For large values of $x$ the behaviour is the opposite. As $x \to \infty$, $\frac{1}{x}$ gets as close to 0 as you like (for example, $\frac{1}{1\,000\,000} = 0.000\,001$) without ever reaching it. The same happens as $x \to -\infty$, where the values are small and negative. Far from the origin, the curve runs alongside the $x$-axis.

A line that a curve approaches in this way is an asymptote. Asymptotes are not part of the graph, so they are drawn as dashed lines.

The graph of $y = \frac{1}{x}$ therefore has the vertical asymptote $x = 0$ (the $y$-axis) and the horizontal asymptote $y = 0$ (the $x$-axis).

Domain, range and symmetry

Since every real number except 0 can be an input, and every real number except 0 is an output (the output $y$ comes from the input $\frac{1}{y}$), the domain and range are

The curve is called a rectangular hyperbola: it has two separate branches, and its two asymptotes are perpendicular.

The graph of $y = \frac{1}{x}$ has three symmetries. It has rotational symmetry of order 2 about the origin, because $\frac{1}{-x} = -\frac{1}{x}$: the point $(x, y)$ is on the curve exactly when $(-x, -y)$ is. It is symmetric in the line $y = x$, because swapping the coordinates of $(2, \frac{1}{2})$ gives $(\frac{1}{2}, 2)$, which is also on the curve. It is also symmetric in the line $y = -x$. The vertices of the hyperbola, the points closest to the origin, are $(1, 1)$ and $(-1, -1)$, where the curve crosses $y = x$.

The reciprocal function is self-inverse

The inverse of a function reverses it (see 2.2), and its graph is the reflection of the original graph in the line $y = x$. Since the graph of $y = \frac{1}{x}$ is its own reflection in $y = x$, the reciprocal function must be its own inverse. The algebra confirms it. Swap $x$ and $y$ in $y = \frac{1}{x}$ and rearrange:

So $f^{-1}(x) = \frac{1}{x} = f(x)$. Applying the function twice returns the original input: $f(f(x)) = \frac{1}{1/x} = x$. Taking the reciprocal of a reciprocal gives back the starting number.

The family y = k/x

Multiplying the reciprocal function by a non-zero constant $k$ gives $y = \frac{k}{x}$. Every member of this family keeps the asymptotes $x = 0$ and $y = 0$, the domain $x \neq 0$ and the range $y \neq 0$. The constant changes where the branches lie:

• If $k > 0$, the branches lie in the first and third quadrants, like $y = \frac{1}{x}$, and $y$ decreases on each branch as $x$ increases.

• If $k < 0$, the branches lie in the second and fourth quadrants, and $y$ increases on each branch as $x$ increases.

• A larger value of $|k|$ moves the branches further from the origin. The vertices are at $x = \pm\sqrt{|k|}$.

Each function in the family is self-inverse, by the same working as in the example: $\frac{k}{\,k/x\,} = x$. The product of the coordinates of every point on $y = \frac{k}{x}$ is the constant $xy = k$, which gives a quick way to find $k$ from one point on the graph.

Rational functions of the form (ax + b)/(cx + d)

What a rational function is

A rational number is a quotient of two integers; a rational function is a quotient of two polynomials.

Examples include $\frac{1}{x}$, $\frac{2x + 3}{x - 1}$ and $\frac{x^2 + 1}{x - 4}$. The functions studied here have a linear numerator and a linear denominator:

The condition $c \neq 0$ makes the denominator depend on $x$; if $c = 0$ the function is just a straight line. The condition $ad - bc \neq 0$ rules out a hidden constant function.

The vertical asymptote

A rational function is undefined where its denominator is zero. For $\frac{ax + b}{cx + d}$, that happens when $cx + d = 0$, that is, at $x = -\frac{d}{c}$. Close to this value, the denominator is a very small number while the numerator is close to a non-zero value, so the quotient is very large in size. For example, with $f(x) = \frac{2x + 3}{x - 1}$:

The values fall without bound just to the left of $x = 1$ and rise without bound just to the right, so the graph has the vertical asymptote $x = 1$.

The numerator must be non-zero at that value for the line to be an asymptote; the condition $ad - bc \neq 0$ guarantees this.

The horizontal asymptote

To see what happens for large $|x|$, divide every term in the numerator and denominator by $x$:

As $x \to \pm\infty$, the terms $\frac{b}{x}$ and $\frac{d}{x}$ both tend to 0, so $f(x) \to \frac{a}{c}$. For large $x$, only the $x$-terms matter, and their ratio is $\frac{ax}{cx} = \frac{a}{c}$.

When the numerator has no $x$-term ($a = 0$), as in $\frac{5}{x + 2}$, the horizontal asymptote is $y = 0$.

Rational functions as transformed reciprocal functions

Any function of this form can be rewritten so that its connection with $\frac{1}{x}$ is visible. For $f(x) = \frac{2x + 3}{x - 1}$, write the numerator as a multiple of the denominator plus a remainder:

In this form the graph is $y = \frac{5}{x}$ translated 1 unit right and 2 units up (translations are studied in 2.11). The asymptotes $x = 0$ and $y = 0$ of $y = \frac{5}{x}$ move to $x = 1$ and $y = 2$, and since $5 > 0$ the branches lie to the upper right and lower left of the point where the asymptotes cross. In general,

so every non-constant function of this form is a hyperbola with the asymptotes found earlier, and its graph has no turning points. This equivalence between two forms of the same function is useful in both directions: the original form gives the intercepts easily, and the transformed form shows which pair of quadrants, relative to the asymptotes, the branches occupy.

Intercepts

The intercepts are found as for any graph.

• $y$-intercept: put $x = 0$. Then $f(0) = \frac{b}{d}$, so the $y$-intercept is $\left(0, \frac{b}{d}\right)$, provided $d \neq 0$. If $d = 0$, the vertical asymptote is the $y$-axis and there is no $y$-intercept.

• $x$-intercept: a fraction is zero only when its numerator is zero, so solve $ax + b = 0$. The $x$-intercept is $\left(-\frac{b}{a}, 0\right)$

Domain and range

The only value of $x$ that cannot be used is the one that makes the denominator zero, and the only value of $y$ that is never reached is the horizontal asymptote. For a non-constant function of this form,

To see why $y = \frac{a}{c}$ is never reached, try to solve $\frac{ax + b}{cx + d} = \frac{a}{c}$. Cross-multiplying gives $acx + bc = acx + ad$, so $bc = ad$, which contradicts $ad - bc \neq 0$. There is no solution.

How to sketch the graph

The method in the example works for every function of this form.

1. Find the vertical asymptote (denominator $= 0$) and the horizontal asymptote (ratio of the $x$-coefficients). Draw both as dashed lines and label them with their equations.

2. Find the $y$-intercept ($x = 0$) and the $x$-intercept (numerator $= 0$), and plot them.

3. Decide which two opposite "corners" formed by the asymptotes contain the branches. The intercepts usually settle this; if not, evaluate $f$ at one value of $x$ on each side of the vertical asymptote.

4. Draw each branch as a smooth curve that gets steadily closer to both asymptotes, passing through the plotted points.

The two branches always sit in opposite corners. If one branch is in the lower-left corner, the other is in the upper-right corner, and if one is in the upper-left, the other is in the lower-right.

Finding a rational function from its features

When the asymptotes or points on a graph are given, the parameters in the equation can be found by matching each feature with its formula.

The inverse of a rational function

Finding the inverse

A non-constant function $f(x) = \frac{ax + b}{cx + d}$ is one-to-one: each horizontal line, apart from the horizontal asymptote, meets its graph exactly once. So it has an inverse function, which is found by the usual method (see 2.5): write $y = f(x)$, swap $x$ and $y$, and make $y$ the subject. The key step is to collect all the terms containing $y$ on one side, then factorise out $y$.

For the general function,

The inverse is another function of the same form. Its vertical asymptote is $x = \frac{a}{c}$ and its horizontal asymptote is $y = -\frac{d}{c}$, so the asymptotes of $f$ and $f^{-1}$ swap roles, just as reflecting in $y = x$ swaps every vertical line with a horizontal one. In particular, the domain of $f^{-1}$ is the range of $f$, $x \neq \frac{a}{c}$, and the range of $f^{-1}$ is the domain of $f$, $y \neq -\frac{d}{c}$.

Self-inverse rational functions

Comparing $f$ with the general inverse shows when a function of this form is its own inverse. The inverse $\frac{-dx + b}{cx - a}$ is the same as $\frac{ax + b}{cx + d}$ exactly when $d = -a$. In that case the vertical asymptote $x = \frac{a}{c}$ and the horizontal asymptote $y = \frac{a}{c}$ cross at a point on the line $y = x$, and the whole graph is symmetric in that line.

Rational functions as models

Inverse proportion

Two quantities are inversely proportional when their product stays the same as they change. If one is doubled, the other is halved; if one is multiplied by 10, the other is divided by 10.

The graph of an inverse proportion is the branch of $y = \frac{k}{x}$ in the first quadrant only, because both quantities are positive in context. The domain of the model is restricted by the situation, so the third-quadrant branch has no meaning.

Boyle's law is an example from physics. For a fixed amount of gas at a constant temperature, the pressure $P$ and the volume $V$ satisfy $PV = k$. Compressing the gas into a smaller volume makes the molecules hit the walls of the container more often, so the pressure rises in proportion.

Travel time and speed

For a journey of fixed distance $d$, the time taken $t$ and the average speed $v$ are inversely proportional, because $vt = d$. The model $t = \frac{d}{v}$ shows a pattern of diminishing returns: each extra 10 km/h of speed saves less time than the one before, because the curve flattens as $v$ increases.

Models of the form (ax + b)/(cx + d)

Rational functions with a non-zero horizontal asymptote model quantities that settle towards a limiting value. The horizontal asymptote then has a meaning in context, often a long-term value that is approached but never reached.

A typical case is a mixing problem. A tank initially holds 100 litres of pure water. Salt water containing 30 grams of salt per litre is pumped in at 5 litres per minute, and nothing flows out. After $t$ minutes the tank contains $150t$ grams of salt in $100 + 5t$ litres, so the concentration in grams per litre is

The horizontal asymptote is $C = 30$: as more and more salt water is added, the concentration in the tank approaches that of the incoming liquid. The vertical asymptote $t = -20$ lies outside the domain, so it plays no part in the model.

Exam-style questions on rational functions

The two questions below show the standard types: a structured non-calculator question on the graph and inverse of a rational function, and a modelling question with a GDC.

Self-check

• To do:

  I can sketch $y = \frac{1}{x}$ and $y = \frac{k}{x}$ for positive and negative $k$, with both asymptotes.

• To do:

  I can explain, graphically and algebraically, why $f(x) = \frac{1}{x}$ is self-inverse.

• To do:

  I can write down the asymptotes of $y = \frac{ax + b}{cx + d}$ and find its intercepts.

• To do:

  I can sketch the graph of $y = \frac{ax + b}{cx + d}$, placing both branches in the correct corners of the asymptotes.

• To do:

  I can state the domain and range of a rational function of this form.

• To do:

  I can find the inverse of $f(x) = \frac{ax + b}{cx + d}$ and state its domain.

• To do:

  I can find unknown constants in a rational function from its asymptotes or a point on its graph.

• To do:

  I can model inverse proportion with $y = \frac{k}{x}$ and interpret asymptotes in context.

1. (a)

   Write down the equations of the vertical asymptotes of the graph of $y = f(x)$.

   [2]2 marks

   Markscheme (a)

   $x = \frac{\pi}{4}$ A1

   $x = \frac{3\pi}{4}$ A1

2. (b)

   Find the range of $f$ for $\frac{\pi}{4} < x < \frac{3\pi}{4}$

3. (c)

   Solve the equation $f(x) = x$.

   [2]2 marks

   Markscheme (c)

   attempt to find the intersection of $y = 3 - \sec 2x$ and $y = x$

Since $(g \circ g)(x) = x$ for every $x \neq 0$, applying $g$ twice undoes itself, so $g^{-1}(x) = g(x)$ and $g$ is self-inverse.

(b) Because $g^{-1} = g$, you can write down $g^{-1}(3) = g(3) = \frac{6}{3} = 2$. As a check, $g(2) = 3$, so $g$ does send 2 to 3, and the inverse sends 3 back to 2.

The equation is $y = -\frac{12}{x}$, so the branches lie in the second and fourth quadrants.

(b) Since $k < 0$, the graph has no points in the first or third quadrant, so it cannot meet $y = x$ (which lies in those quadrants). For $y = -x$, substitute into the equation:

When $x = 2\sqrt{3}$, $y = -2\sqrt{3}$; when $x = -2\sqrt{3}$, $y = 2\sqrt{3}$. The points are $(2\sqrt{3}, -2\sqrt{3})$ and $(-2\sqrt{3}, 2\sqrt{3})$, the vertices of this hyperbola.

(b) When $x = 0$, $f(0) = \frac{3}{-1} = -3$, so the $y$-intercept is $(0, -3)$. The graph meets the $x$-axis when the numerator is zero: $2x + 3 = 0$, so $x = -1.5$. The $x$-intercept is $(-1.5, 0)$.

(c) Draw the asymptotes as dashed lines and plot the intercepts. Both intercepts lie to the left of $x = 1$ and below $y = 2$, so they are on the lower-left branch. Since $f(x) = 2 + \frac{5}{x - 1}$ with $5 > 0$, the other branch lies to the upper right of the point $(1, 2)$. A check value confirms it: $f(2) = \frac{7}{1} = 7$.

(d) Every value except the horizontal asymptote is reached, so the range is $y \in \mathbb{R}$, $y \neq 2$.

Vertical asymptote: $2x + 4 = 0$ gives $x = -2$.

Horizontal asymptote: $y = \frac{a}{c} = -\frac{1}{2}$.

$y$-intercept: $g(0) = \frac{3}{4}$, giving $\left(0, \frac{3}{4}\right)$.

$x$-intercept: $3 - x = 0$ gives $(3, 0)$.

Both intercepts lie to the right of $x = -2$ and above $y = -\frac{1}{2}$, so one branch is in the upper-right corner. The other branch must be in the lower-left corner. A check value: $g(-3) = \frac{6}{-2} = -3$, which is below $-\frac{1}{2}$, as expected.

Domain: $x \in \mathbb{R}$, $x \neq -2$. Range: $y \in \mathbb{R}$, $y \neq -\frac{1}{2}$.

The horizontal asymptote is the ratio of the $x$-coefficients, $\frac{a}{1} = a$. Since it is $y = 3$, $a = 3$.

So $f(x) = \frac{3x + 5}{x + 2}$. Check that this is not a constant function: $ad - bc = 3(2) - 5(1) = 1 \neq 0$.

(b) The graph crosses the $x$-axis when the numerator is zero: $3x + 5 = 0$, so $x = -\frac{5}{3}$. The point is $\left(-\frac{5}{3}, 0\right)$.

Multiply both sides by $y - 1$, then collect the terms in $y$ on the left:

So $f^{-1}(x) = \frac{x + 3}{x - 2}$.

(b) The domain of $f^{-1}$ is the range of $f$, so $x \in \mathbb{R}$, $x \neq 2$; this matches the zero of the new denominator. The range of $f^{-1}$ is the domain of $f$: $y \in \mathbb{R}$, $y \neq 1$, which is the horizontal asymptote of $y = \frac{x + 3}{x - 2}$.

(c) $f(2) = \frac{7}{1} = 7$ and $f^{-1}(7) = \frac{10}{5} = 2$. Applying $f$ and then $f^{-1}$ returns the starting value, as it should.

So $h^{-1}(x) = \frac{4x + 1}{2x - 4} = h(x)$, and $h$ is self-inverse. As a check, the asymptotes $x = 2$ and $y = \frac{4}{2} = 2$ meet at $(2, 2)$, which lies on $y = x$.

(b) $P = \frac{600}{1.5} = 400$ kPa. The volume has been divided by 4, so the pressure is multiplied by 4.

(c) $V = \frac{600}{P} = \frac{600}{250} = 2.4$ litres.

(d) As $V \to \infty$, $P \to 0$: the horizontal asymptote $P = 0$ means that spreading the gas out lowers the pressure towards zero without ever reaching it. As $V \to 0^+$, $P \to \infty$, which is the vertical asymptote $V = 0$. A volume of 0 is not possible: the model predicts an infinite pressure, and in reality the gas would turn into a liquid long before. The model is valid only for $V > 0$, and in practice only over a moderate range of pressures.

The concentration reaches 20 grams per litre after 40 minutes.

(c) $C = 30$ is the horizontal asymptote, which is not in the range. Algebraically, $\frac{30t}{20 + t} = 30$ gives $30t = 600 + 30t$, so $0 = 600$, which is impossible. The tank always contains some of the original pure water, so the mixture is always slightly weaker than the incoming salt water. In practice the tank would also overflow at some point, which limits the domain further.

(b) $f(0) = \frac{2}{2} = 1$, so the $y$-intercept is $(0, 1)$. Setting the numerator to zero, $3x + 2 = 0$ gives $x = -\frac{2}{3}$, so the $x$-intercept is $\left(-\frac{2}{3}, 0\right)$.

(c) Both intercepts lie to the right of $x = -2$ and below $y = 3$, so one branch is in the lower-right corner of the asymptotes, and the other branch is in the upper-left corner. A check point: $f(-3) = \frac{-7}{-1} = 7$.

(d) Swap $x$ and $y$ in $y = \frac{3x + 2}{x + 2}$ and rearrange:

The domain of $f^{-1}$ is the range of $f$: $x \in \mathbb{R}$, $x \neq 3$.

(e) Multiply both sides of $\frac{3x + 2}{x + 2} = x$ by $x + 2$:

So $x = 2$ or $x = -1$; neither is the excluded value $-2$. Points where $f(x) = x$ lie on $y = x$, and the reflection of such a point in $y = x$ is itself, so these points lie on the graph of $f^{-1}$ as well. The graphs meet at $(2, 2)$ and $(-1, -1)$. (Solving $f(x) = f^{-1}(x)$ directly leads to $5x^2 - 5x - 10 = 0$, the same equation, so there are no other meeting points.)

(b) Solve $\frac{12\,000 + 35n}{n} = 50$, either with the GDC (intersection of $y = A(x)$ and $y = 50$) or algebraically: $12\,000 + 35n = 50n$, so $15n = 12\,000$ and $n = 800$ lamps.

(c) Writing $A(n) = 35 + \frac{12\,000}{n}$ shows the horizontal asymptote $A = 35$. As production increases, the fixed costs are shared between more and more lamps, so the average cost per lamp approaches the material cost of USD 35 but never falls to it.

(d) $A(0)$ would involve dividing by zero: with no lamps made, there is no average cost per lamp. The line $n = 0$ is a vertical asymptote of the graph.

[2]2 marks

$x = 0.573$ $(0.573052\ldots)$ A1

Note: Award (M1)A0 if any additional solutions are given.