Lesson

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

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

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

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

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

A hyperbola can be scaled by changing the value of the numerator. So to expand the hyperbola $y=\frac{1}{x}$`y`=1`x` by a scale factor of $a$`a` we get $y=\frac{a}{x}$`y`=`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.

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

Summary

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

Hyperbolas can have $0$0 or $1$1 $x$`x`-intercepts and can have $0$0 or $1$1$y$`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.

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

- Vertically translated by $k$
`k`units: $y=\frac{1}{x}+k$`y`=1`x`+`k` - Horizontally translated by $h$
`h`units: $y=\frac{1}{x-h}$`y`=1`x`−`h` - Scaled by a scale factor of $a$
`a`: $y=\frac{a}{x}$`y`=`a``x` - Reflected about the $y$
`y`-axis: $y=-\frac{1}{x}$`y`=−1`x`

Consider the function $y=\frac{1}{x}$`y`=1`x` which is defined for all real values of $x$`x` except $0$0.

Complete the following table of values.

$x$ `x`$-2$−2 $-1$−1 $-\frac{1}{2}$−12 $-\frac{1}{4}$−14 $\frac{1}{4}$14 $\frac{1}{2}$12 $1$1 $2$2 $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Plot the points in the table of values.

Loading Graph...Hence draw the curve.

Loading Graph...In which quadrants does the graph lie?

$3$3

A$2$2

B$1$1

C$4$4

D$3$3

A$2$2

B$1$1

C$4$4

D

Consider the function $y=-\frac{1}{2x}$`y`=−12`x`

Complete the following table of values.

$x$ `x`$-3$−3 $-2$−2 $-1$−1 $1$1 $2$2 $3$3 $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Sketch the graph.

Loading Graph...In which quadrants does the graph lie?

$2$2

A$1$1

B$3$3

C$4$4

D$2$2

A$1$1

B$3$3

C$4$4

D

Consider the equation $f\left(x\right)=\frac{3}{x}$`f`(`x`)=3`x`.

Sketch a graph of the function on the axes below:

Loading Graph...Which of the following statements about the symmetry of the graph is true?

The graph is symmetric about the $x$

`x`-axisAThe graph is symmetric about the $y$

`y`-axis.BThe graph has no symmetry.

CThe graph is rotationally symmetric about the origin.

DThe graph is symmetric about the $x$

`x`-axisAThe graph is symmetric about the $y$

`y`-axis.BThe graph has no symmetry.

CThe graph is rotationally symmetric about the origin.

DFind an expression for $f\left(-x\right)$

`f`(−`x`):

Describe, interpret and sketch parabolas, hyperbolas, circles and exponential functions and their transformations.