The Hyperbola y = a/(x + p) + q

Q: How do I find the asymptotes of a hyperbola?

The vertical asymptote is the $x$-value that makes the denominator 0. The horizontal asymptote is the constant added at the end, $q$.

The Grade 11 hyperbola $y = \frac{a}{x + p} + q$ can be shifted left, right, up and down. Its asymptotes and axes of symmetry are the key features, and finding the equation from a sketch is a standard question.

Asymptotes

For $y = \frac{a}{x + p} + q$ :

Vertical asymptote: $x = - p$ (the value that makes the denominator 0)
Horizontal asymptote: $y = q$

Domain: $x \neq = - p$ . Range: $y \neq = q$ .

Example. $y = \frac{4}{x + 1} - 2$ has asymptotes $x = - 1$ and $y = - 2$ .

The sign of $a$ decides which quadrants (relative to the asymptotes) the graph lies in: $a > 0$ gives top-right and bottom-left; $a < 0$ gives top-left and bottom-right.

Intercepts and sketching

$y$ -intercept: let $x = 0$ .
$x$ -intercept: let $y = 0$ and solve.

Example. $y = \frac{4}{x + 1} - 2$

$y$ -intercept: $\frac{4}{1} - 2 = 2$ , so $(0; 2)$
$x$ -intercept: $\frac{4}{x + 1} = 2 \Rightarrow x + 1 = 2 \Rightarrow x = 1$ , so $(1; 0)$

To sketch: draw the asymptotes as dotted lines, plot the intercepts, then draw the two branches so they approach the asymptotes without touching.

Axes of symmetry

A hyperbola has two axes of symmetry. Both pass through the point where the asymptotes meet, $(- p; q)$ , with gradients $1$ and $- 1$ :

$y = (x + p) + q and y = - (x + p) + q$

Example. For $y = \frac{4}{x + 1} - 2$ , the asymptotes meet at $(- 1; - 2)$ :

$y = x + c$ : $- 2 = - 1 + c$ , so $y = x - 1$
$y = - x + c$ : $- 2 = 1 + c$ , so $y = - x - 3$

Finding the equation from a graph

Read $p$ and $q$ from the asymptotes.
Substitute one point on the graph to find $a$ .

Example. Asymptotes $x = 2$ and $y = 3$ , passing through $(4; 5)$ :

$y = \frac{a}{x - 2} + 3 \Rightarrow 5 = \frac{a}{2} + 3 \Rightarrow a = 4$

$y = \frac{4}{x - 2} + 3$

Key terms

Hyperbola: The graph of $y = \frac{a}{x + p} + q$ , with two separate branches.
Asymptote: A line the graph approaches but never touches.
Domain: All $x$ -values for which the function is defined.

Common mistakes

Writing the vertical asymptote of $\frac{4}{x + 1}$ as $x = 1$ .
Drawing branches that cross an asymptote.
Using the wrong point for the axes of symmetry; it must be the intersection of the asymptotes.
Forgetting to multiply both sides by $(x + p)$ carefully when finding the $x$ -intercept.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

For $y = \frac{4}{x + 1} - 2$ , write down:

(a) the equations of the asymptotes (2)

(b) the $y$ -intercept (1)

Worked solution

Hidden - try the question first

Solution

(a) $x = - 1$
$y = - 2$
(b) $y = \frac{4}{1} - 2 = 2$

Question 2

5 marks total

routine

For $y = \frac{4}{x + 1} - 2$ :

(a) Determine the $x$ -intercept. (2)

(b) Sketch the graph, showing asymptotes and intercepts. (3)

Worked solution

Hidden - try the question first

Solution

(a) $x = - 1$
$x = 1$
(b) Asymptotes drawn as dotted lines at $x = - 1$ and $y = - 2$
Intercepts $(0; 2)$ and $(1; 0)$ on the right branch
Left branch in the bottom-left region, both branches approaching the asymptotes

Question 3

6 marks total

complex

For $y = \frac{4}{x + 1} - 2$ :

(a) Determine the equations of both axes of symmetry. (4)

(b) Write down the domain and range. (2)

Worked solution

Hidden - try the question first

Solution

(a) Asymptotes meet at $x = - 1$
$y = x + c$ : $- 2 = - 1 + c \Rightarrow y = x - 1$
$y = - x + c$ : $- 2 = 1 + c$
$y = - x - 3$
(b) Domain: $x \neq = - 1$
Range: $y \neq = - 2$

Question 4

8 marks total

problem-solving

A hyperbola $y = \frac{a}{x + p} + q$ has asymptotes $x = 2$ and $y = 3$ , and passes through $(4; 5)$ .

(a) Determine the values of $a$ , $p$ and $q$ . (3)

(b) Determine the $x$ -intercept. (3)

Worked solution

Hidden - try the question first

Solution

(a) $x = - 1$ and $q = 3$
$5 = \frac{a}{4 - 2} + 3$
$a = 4$
(b) $\frac{4}{x - 2} = - 3$
$4 = - 3 x + 6$
$x = \frac{2}{3}$
(c) $a > 0$ , so the graph is above $y = 3$ to the right of the vertical asymptote
$x > 2$

FAQs

How do I find the asymptotes of a hyperbola?

The vertical asymptote is the $x$ -value that makes the denominator 0. The horizontal asymptote is the constant added at the end, $q$ .

Why does a hyperbola have two axes of symmetry?

The two branches are mirror images in two diagonal lines through the centre where the asymptotes meet.

Is the hyperbola examined in Grade 12 too?

Yes. Grade 12 function questions often include a hyperbola alongside other graphs.

Asymptotes

Intercepts and sketching

Axes of symmetry

Finding the equation from a graph

Key terms

Common mistakes

Practice questions

FAQs

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