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8 Nonlinear relationships
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Stage 5.1-3

8.09 Hyperbolas

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Worksheet
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Lesson

Features of hyperbolas

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Graphs of equations of the form \\ y=\dfrac{a}{x-h}+k (where a, and h and k are any number) are called hyperbolas.

The graph shows the hyperbola y=\dfrac{1}{x}.

Hyperbolas can have either 0 or 1 \, x-intercepts. This is the point on the graph which touches the x-axis. We can find this by setting y=0 and finding the value of x. If the x-value is undefined, there is no x-intercept. For example, there is no x-intercept of y=\dfrac{1}{x}.

Similarly, hyperbolas can have either 0 or 1 \, y-intercept. This is the point on the graph which touches the y-axis. We can find this by setting x=0 and finding the value of y. If the y-value is undefined, there is no y-intercept. For example, there is no y-intercept of y=\dfrac{1}{x}.

Hyperbolas have a vertical asymptote which is the vertical line which the graph approaches but does not touch. For example, the vertical asymptote of y=\dfrac{1}{x} is x=0.

Hyperbolas also have a horizontal asymptote which is the horizontal line which the graph approaches but does not touch. For example, the horizontal asymptote of y=\dfrac{1}{x} is y=0..

Examples

Example 1

Consider the function y = - \dfrac{1}{4 x}.

a

Complete the following table of values:

x-3-2-1123
y
Worked Solution
Create a strategy

Substitute the values from the table into the equation.

Apply the idea

For x=-3:

\displaystyle y\displaystyle =\displaystyle -\dfrac{1}{4x}
\displaystyle =\displaystyle -\dfrac{1}{4(-3)}Substitute x=-3
\displaystyle =\displaystyle \dfrac{1}{12}Evaluate

Similarly, by substituting the remaining x-values into y=-\dfrac{1}{4x}, we get:

x-3-2-1123
y\dfrac{1}{12}\dfrac{1}{8}\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{12}
b

Sketch the graph.

Worked Solution
Create a strategy

Plot the points in the table of values and draw the curve passing through each plotted point.

Apply the idea
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The ordered pairs of points to be plotted on the coordinate plane are (-3,\dfrac{1}{12}), (-2,\dfrac{1}{8}), (-1,\dfrac{1}{4}), (1,-\dfrac{1}{4}), (2,-\dfrac{1}{8}),

and (3,-\dfrac{1}{12}), which are plotted on the graph.

The curve of the hyperbola y=-\dfrac{1}{4x} must pass through each of the plotted points and approaches the horizontal asymptote y=0 and the horizontal asymptote x=0.

c

In which quadrants does the graph lie?

Worked Solution
Apply the idea

From the graph in part(b), we see that the hypeybola lies in quadrant 2 and quadrant 4.

Idea summary

The graph of an equation of the form y=\dfrac{a}{x-h}+k is a hyperbola.

Hyperbolas can have 0 or 1 x-intercepts and can have 0 or 1 y-intercepts, depending on the solutions to the equation.

Hyperbola have a vertical asymptote which is the vertical line that the graph approaches but does not intersect and a horizontal asymptote which is the horizontal line that the graph approaches but does not intersect.

Transformations of hyperbolas

A hyperbola can be vertically translated by increasing or decreasing the y-values by a constant number. So to translate y=\dfrac{1}{x} up by k units gives us y=\dfrac{1}{x} + k.

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This graph shows y=\dfrac{1}{x} translated vertically up by 2 to get y=\dfrac{1}{x}+2, and down by 2 to get y=\dfrac{1}{x}-2.

Similarly, a hyperbola can be horizontally translated by increasing or decreasing the x-values by a constant number. However, the x-value together with the translation must both be in the denominator. That is, to translate y=\dfrac{1}{x} to the left by h units we get y=\dfrac{1}{x+h}.

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This graph shows y=\dfrac{1}{x} translated horizontally left by 2 to get y=\dfrac{1}{x+2} and right by 2 to get y=\dfrac{1}{x-2}.

A hyperbola can be scaled by changing the value of the numerator. So to expand the hyperbola y=\dfrac{1}{x} by a scale factor of a we get y=\dfrac{a}{x}. We can compress a hyperbola by dividing by the scale factor instead. Note that for hyperbolas, vertically scaling is equivalent to horizontally scaling.

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This graph shows y=\dfrac{1}{x} vertically expanded by a scale factor of 2 to get y=\dfrac{2}{x} and compressed by a scale factor of 2 to get y=\dfrac{1}{2x}.

We can reflect a hyperbola about either axis by taking the negative of the y-values. So to reflect y=\dfrac{1}{x} about the x-axis gives us y=-\dfrac{1}{x}. Note that for hyperbolas, vertically reflecting is equivalent to horizontally reflecting.

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This graph shows y=\dfrac{1}{x} reflected about the y-axis which gives us y=-\dfrac{1}{x}.

Exploration

The following applet demonstrates how a scale factor affects the shape of a hyperbola. Play with the applet below by dragging the sliders.

Loading interactive...

As the scale factor a increases in size, the hyperbola moves away from the axes. If a is negative, the hyperbola is reflected across the x-axis.

Examples

Example 2

Consider the equation f(x) = \dfrac{3}{x}.

a

Sketch a graph of the function.

Worked Solution
Create a strategy

Consider the transformation that has occured on y=\dfrac{1}{x}.

Apply the idea

To get the graph of y=\dfrac{3}{x} we would need to vertically scale the hyperbola y=\dfrac{1}{x} by 3. This means that the graph of y=\dfrac{3}{x} would have the same asymptotes and lie in the same quadrants as the hyperbola y=\dfrac{1}{x}.

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The point (1,1) on the hyperbola y=\dfrac{1}{x} would be vertically scaled to the point (1,3). \, \,(-1,-1) would be scaled to the (-1,-3). \, \,\left(2,\dfrac{1}{2}\right) would be scaled to \left(2,\dfrac{3}{2}\right).

By plotting these points and drawing the curve through them we get the hyperbola shown.

b

Which of the following statements about the symmetry of the graph is true?

A
The graph is symmetric about the x-axis.
B
The graph is symmetric about the y-axis.
C
The graph has no symmetry.
D
The graph is rotationally symmetric about the origin.
Worked Solution
Create a strategy

Visualise reflecting the graph about the axes and rotating the graph about the origin.

Apply the idea

Reflecting the graph along the x-axis, or the y-axis, will move the hyperbola to the wrong quadrants.

If we rotate the graph 180\degree about the origin, it will match up to itself. So the graph is rotationally symmetric about the origin.

Option D is the correct answer.

c

Find an expression for f(-x).

Worked Solution
Create a strategy

Replace x with -x in the function.

Apply the idea
\displaystyle f(x)\displaystyle =\displaystyle \dfrac{3}{x}
\displaystyle f(-x)\displaystyle =\displaystyle \dfrac{3}{-x}Substitute -x
\displaystyle f(-x)\displaystyle =\displaystyle -\dfrac{3}{x}Simplify
Idea summary

Hyperbolas can be transformed in the following ways (starting with the hyperbola defined by y=\dfrac{1}{x}):

  • Reflected about the y-axis: y=-\dfrac{1}{x}

  • Vertically translated by k units: y=\dfrac{1}{x} + k

  • Horizontally translated by h units: y=\dfrac{1}{x-h}

  • Scaled by a scale factor of a: y=\dfrac{a}{x}

Outcomes

MA5.3-9NA

sketches and interprets a variety of non-linear relationships

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