Hyperbolas and their asymptotes

What is it about?

Two quantities are inversely proportional if their product is always the same: if one doubles, the other halves. For a fixed number $a$ :

x \cdot y = a \quad\Longleftrightarrow\quad y = \frac{a}{x} \qquad (x \neq 0)

In a picture: all rectangles between the axes and a point $P(x \mid y)$ on the graph have the same area $|a|$ – a narrow tall rectangle just like a wide flat one. The graph is called a hyperbola; the simplest one is the basic function $y = \frac{1}{x}$ .

Stretching by the factor $a$ and then shifting by $b$ in the $x$ -direction and by $c$ in the $y$ -direction gives

f(x) = \frac{a}{x - b} + c

Vertical asymptote $x = b$ : there the denominator is zero and $f$ is not defined (pole). As $x$ gets closer and closer to $b$ , the absolute values of $f(x)$ grow without bound.
Horizontal asymptote $y = c$ : for $x \to +\infty$ and $x \to -\infty$ , $\frac{a}{x-b}$ gets arbitrarily small; the graph approaches the line $y = c$ but never reaches it.
Domain $D = \mathbb{R} \setminus \{b\}$ , range $W = \mathbb{R} \setminus \{c\}$ .
Sign of $a$ : relative to the cross formed by the asymptotes (intersection $Z(b \mid c)$ ), the two branches lie in quadrants I and III for $a > 0$ (decreasing on both intervals) and in quadrants II and IV for $a < 0$ (increasing). The larger $|a|$ , the farther the branches are from the asymptotes.
The inverse proportion still holds after shifting – measured from the asymptotes: $(x - b) \cdot (y - c) = a$ .

Intersections with the axes:

f(x) = 0 \;\Longleftrightarrow\; x = b - \frac{a}{c} \quad (c \neq 0), \qquad f(0) = c - \frac{a}{b} \quad (b \neq 0)

For $c = 0$ the $x$ -axis itself is the asymptote – then there is no zero; for $b = 0$ there is no $y$ -intercept.

Try it

Drag the orange point $P$ along the graph. The green rectangle changes its shape, but its area stays $4$ . Look at the table of values: what happens to $y$ when $x$ doubles? Load example
Make $a$ negative. Which quadrants do the branches lie in now? What does that mean for monotonicity? Load example
Click First stretch, then shift: step by step the hyperbola emerges from the basic function $\frac{1}{x}$ . Load example
Drag the asymptote cross $Z$ . How are $D$ and $W$ related to $Z$ ?
Choose “x → ±∞” as table of values and let $P$ run outwards. By how much does the distance to the horizontal asymptote shrink when $x - b$ becomes ten times as large? Load example
Let $P$ run to the pole. What happens to the function values? Load example
Drag the green point $Q(b + 1 \mid c + a)$ : it lies $1$ to the right of $Z$ and is shifted by $a$ (so below $Z$ for $a < 0$ ). The rectangle of width $1$ has the area $|a|$ . If $P$ is right there, $Q$ jumps to the other side: $Q(b - 1 \mid c - a)$ .

Tasks

Task 1: Inversely proportional?

A rectangle is to have an area of $12\,\text{cm}^2$ . Express the length $y$ in terms of the width $x$ and calculate $y$ for $x = 1.5$ ; $3$ ; $6$ .

Show solution

$x \cdot y = 12$ , so $y = \frac{12}{x}$ : $y = 8$ ; $4$ ; $2$ (each in cm). Double the width – half the length. The graph is the branch in quadrant I of the hyperbola $y = \frac{12}{x}$ (only $x > 0$ makes sense). The simulation goes up to $a = 8$ , but it shows the principle just the same: Show

Task 2: Read off the equation

Load task Read off the asymptotes in the picture (not from the sliders) and find $a$ using a point on the graph. State the equation, $D$ and $W$ .

Show solution

Asymptotes $x = 2$ and $y = 1$ , so $b = 2$ , $c = 1$ . The point $P(3 \mid -1)$ lies on the graph: $(3 - 2) \cdot (-1 - 1) = -2 = a$ . So $f(x) = \frac{-2}{x - 2} + 1$ , $D = \mathbb{R} \setminus \{2\}$ , $W = \mathbb{R} \setminus \{1\}$ . The branches lie in quadrants II and IV because $a < 0$ . Show in the simulation

Task 3: Zero and y-intercept

Find the points where the graph of $f(x) = \frac{3}{x - 2} + 1$ meets the coordinate axes.

Show solution

Zero: $\frac{3}{x-2} = -1 \Rightarrow x - 2 = -3 \Rightarrow x = -1$ , so $N(-1 \mid 0)$ . $y$ -axis: $f(0) = \frac{3}{-2} + 1 = -0.5$ , so $S_y(0 \mid -0.5)$ . Show in the simulation

Task 4: Approaching the asymptote

For $f(x) = \frac{2}{x + 1} + 1.5$ : from which $x$ on do the function values differ from $1.5$ by less than $0.01$ (for $x > -1$ )?

Show solution

$\frac{2}{x+1} < 0.01 \Leftrightarrow x + 1 > 200 \Leftrightarrow x > 199$ . The table of values shows: at $x - b = 100$ the distance is still $0.02$ , at $1000$ only $0.002$ . Show in the simulation

Task 5: Hyperbola through two points

The hyperbola $f(x) = \frac{a}{x} + c$ passes through $A(1 \mid 5)$ and $B(4 \mid 2)$ . Find $a$ and $c$ .

Show solution

$a + c = 5$ and $\frac{a}{4} + c = 2$ . Subtracting gives $\frac{3}{4}a = 3$ , so $a = 4$ and $c = 1$ : $f(x) = \frac{4}{x} + 1$ . Show in the simulation

Ideas for the classroom

Use: About grade 8, elementary rational functions: start with inverse proportion (rectangles of equal area), then the effect of the parameters, domain and range, asymptotes. The form $\frac{a}{x - b} + c$ corresponds to the vertex form of a parabola – the animation “First stretch, then shift” builds on this.
Limits: approaching the asymptotes is treated only informally with tables of values (“gets arbitrarily close but never reaches it”); $x \to \infty$ is a way of speaking here, not a formal limit.
Common stumbling blocks:
- “The graph never crosses the asymptote.” – True for these hyperbolas; in general (later: rational functions) a graph may well cross its horizontal asymptote.
- The sign of $b$ : for $\frac{a}{x + 2}$ the pole is at $x = -2$ .
- “The two branches are two functions.” – It is one function with a gap in its domain; the branches are point-symmetric about $Z(b \mid c)$ .
- The pole is mistaken for a zero because the denominator is zero there.
Task mode: with info=0 and Share → hide results, the key features, table of values and labels are hidden; asymptotes and graph stay visible and can be read off.