IB Applications and Interpretation SL · Functions · Section 6 of 7

Integer power functions and vertical and horizontal asymptotes

3 worked examples · By Álvaro López

Key idea

Functions with integer powers of $x$ form a broad family: cubics, functions with $1/ x$ , with $1/ x^{2}$ … Their graphs are analysed with the GDC, and what the exam asks about most often is the local maxima and minima, the intercepts with the axes and the asymptotes, which show how the function behaves close to a forbidden value or very far from the origin.

Positive powers: the cubic function

A cubic function is a polynomial function whose highest power of $x$ is $3$ : $f (x) = a x^{3} + b x^{2} + c x + d, a \neq = 0$ Its graph has an S-shape. If $a > 0$ it rises to the right and falls to the left; if $a < 0$ , the other way round. It can cut the $x$ -axis at one, two or three points, and it can have a local maximum and a local minimum, or neither. The $y$ -intercept is always $(0, d)$ .

In the exam you are not expected to differentiate or factorise a complicated cubic: it is sketched on the GDC and the zeros, the local maximum and the local minimum are read off. If the function is already factorised, you can see the zeros directly.

Negative powers and asymptotes

When $x$ appears in the denominator, the function does not exist for some values. The simplest such function is $f (x) = \frac{1}{x}, x \neq = 0$ If you approach $0$ from the right, $1/ x$ becomes huge and positive. If you approach it from the left, it becomes huge and negative. If $x$ becomes very large, positive or negative, $1/ x$ approaches $0$ .

For $f (x) = \frac{a}{x} + b$ , with $a \neq = 0$ : the vertical asymptote is $x = 0$ and the horizontal one is $y = b$ . If $a > 0$ the graph has two branches, one in the first quadrant relative to the asymptotes and the other in the third. If $a < 0$ , the branches lie in the other two. For $f (x) = \frac{a}{x ^{2}} + b$ , both branches lie above the horizontal asymptote if $a > 0$ , because $x^{2}$ is never negative.

Graph of f(x) = 6/x + 2 with vertical asymptote x = 0, horizontal asymptote y = 2 and x-intercept at (−3; 0)

Worked examples

Example 1 Basic

Asymptotes of a function with 1/x

Let $f (x) = \frac{6}{x} + 2$ , with $x \neq = 0$ .

a) Write down the equation of the vertical asymptote and give the value of $x$ .

b) Write down the equation of the horizontal asymptote and give its value $y$ .

c) Calculate $f (3)$ .

d) Find the value of $x$ for which $f (x) = 0$ .

Solution

a) The function is not defined at $x = 0$ , and as $x$ approaches $0$ the quotient $6/ x$ becomes very large in absolute value. The vertical asymptote is $x = 0$ .

b) As $x$ becomes very large, $6/ x$ approaches $0$ and $f (x)$ approaches $2$ . The horizontal asymptote is $y = 2$ .

c) $f (3) = \frac{6}{3} + 2 = 4$ .

d) Solve $\frac{6}{x} + 2 = 0$ : $\frac{6}{x} = - 2 \Rightarrow x = - 3$

Answer: a) $x = 0$ ; b) $y = 2$ ; c) $4$ ; d) $x = - 3$ .

Example 2 Medium

A cubic with the GDC

Let $f (x) = x^{3} - 3 x^{2} - 9 x + 27$ .

a) Using the GDC, find the values of $x$ at which the graph cuts the $x$ -axis.

b) Find the coordinates of the local maximum and the local minimum.

c) Find the values of $k$ for which the equation $f (x) = k$ has exactly three solutions. Give the two ends of the open interval, from smaller to larger.

Solution

a) Draw the graph on the GDC and find the zeros. We obtain $x = - 3$ and $x = 3$ . This can be checked by factorising: $f (x) = (x - 3)^{2} (x + 3)$ , with $x = 3$ as a double root. That is why the graph touches the $x$ -axis at $x = 3$ without crossing it.

b) Using the GDC's maximum and minimum function, the local maximum is at $(- 1, 32)$ and the local minimum is at $(3, 0)$ . Check: $f (- 1) = - 1 - 3 + 9 + 27 = 32$ and $f (3) = 27 - 27 - 27 + 27 = 0$ .

c) The equation $f (x) = k$ is solved by finding the intersection points of the graph with the horizontal line $y = k$ . The line cuts the graph three times when it lies between the local minimum and the local maximum, that is, when $0 < k < 32$ . For the values $k = 0$ and $k = 32$ the line touches the graph at a point of tangency and there are only two solutions.

Answer: a) $x = - 3$ and $x = 3$ ; b) local maximum $(- 1, 32)$ and local minimum $(3, 0)$ ; c) $0 < k < 32$ .

Example 3 Hard

Cost per person of a coach trip

A school hires a coach for a trip. The hire cost is fixed and is shared among the $n$ students who go. In addition, each student pays for their own food. The cost per student, in euros, is $C (n) = 12 + \frac{180}{n}, n \geq 1$

a) Find the cost per student if $10$ students go.

b) Find how many students must go for the cost per student to be $18$ euros.

c) Write down the equation of the horizontal asymptote of $C$ and explain what it means in this context.

d) Find the smallest number of students for which the cost per student is less than $13$ euros.

Solution

a) $C (10) = 12 + \frac{180}{10} = 12 + 18 = 30$ . Each student pays $30$ euros.

b) Solve $C (n) = 18$ : $12 + \frac{180}{n} = 18 \Rightarrow \frac{180}{n} = 6 \Rightarrow n = 30$ $30$ students must go.

c) As $n$ becomes very large, $\frac{180}{n}$ approaches $0$ and $C (n)$ approaches $12$ . The horizontal asymptote is $C = 12$ . It means that, however many people go, the cost per student never drops below $12$ euros: that is the fixed cost for each one (the food), and the cost of the coach is spread more and more thinly.

d) Set up $C (n) < 13$ : $12 + \frac{180}{n} < 13 \Rightarrow \frac{180}{n} < 1 \Rightarrow n > 180$ As $n$ is an integer, the smallest value is $n = 181$ . It can also be seen on the GDC: draw $C$ and look for where the curve lies below the line $y = 13$ .

In the exam: when you interpret an asymptote, name the quantity and say what happens in context ("the cost per student approaches $12$ euros"), not just the equation. In a context with an integer variable, the final answer must be a valid integer.

Common mistakes

Summary

A cubic $f (x) = a x^{3} + b x^{2} + c x + d$ can have a local maximum and a local minimum; these are located with the GDC.
A vertical asymptote $x = a$ appears when the function does not exist at $a$ and grows without limit near it.
A horizontal asymptote $y = b$ describes the value the function approaches when $x$ is very large or very negative.
For $f (x) = \frac{a}{x} + b$ , the asymptotes are $x = 0$ and $y = b$ .
Solving $f (x) = k$ is equivalent to intersecting the graph with the horizontal line $y = k$ ; in a context, always interpret the asymptotes in words.

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