Trigonometric Graphs

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Transformations of tangent curves and equations

Lesson

We already know that transformations to curves, graphs or equations mean that we are doing one of four things:

**horizontal translation**- shifting the graph horizontally (in cyclic curves we call this a phase shift)**vertical translation**- shifting the curve vertically**reflection**- reflecting the curve in the $y$`y`-axis**dilation**- changing the dilation of the curve (in cyclic curves this effect typically changes the amplitude)

Before we move into the nuts and bolts about how these are reflected in the equations of a $\tan$`t``a``n` function let's get an idea by exploring this applet.

- Can you make the graph
**translate horizontally**and can you see how this is represented in the equation? - Can you make the graph
**translate vertically**and can you see how this is represented in the equation? - Can you
**reflect**the graph in the $y$`y`-axis, and see how this is represented in the equation? - Can you change the
**vertical dilation**, and see how this is represented in the equation? - Change the
**period,**and see how this is represented in the equation?

The general form of the tan functions is

$f\left(x\right)=a\tan\left(bx-c\right)+d$`f`(`x`)=`a``t``a``n`(`b``x`−`c`)+`d`

- As tangent functions reach to infinity in both directions of the $y$
`y`-axis, there is no amplitude.

- The vertical dilation (a stretching or shrinking in the same direction as the $y$
`y`-axis) occurs when the value of a is not one. - If $|a|>1$|
`a`|>1, then the graph is stretched - If $|a|<1$|
`a`|<1 then the graph is compressed - Have another look at the applet above now, and change the a value. Can you see the stretching and shrinking?
- The horizontal dilation (a stretching of shrinking in the same direction as the $x$
`x`-axis) occurs when the period is changed, see the next point.

- If $a$
`a`is negative, then there is a reflection. Have a look at the applet above and make a negative, can you see what this does to the curve?

- The period is calculated using $\frac{\pi}{b}$π
`b` - From a graph you can read the value for the period directly by measuring the distance for one complete cycle.
- For a tangent function this occurs between the asymptotes.
- When the period is increased, ie is$>\pi$>π, then the graphs horizontal dilation can be described as a stretch.
- When the period is decrease, ie is $<\pi$<π, then the graphs horizontal dilation can be described as being shrunk or compressed.

- The whole function is shifted by $d$
`d`units. - From an equation you can read the value from the equation directly. If $d>0$
`d`>0 then the graph is translated up, if $d<0$`d`<0 then the graph is translated down.

- The phase shift is the comparable transformation to a horizontal translation. The phase shift is found by calculating $\frac{c}{b}$
`c``b` from the equation

To construct an equation of the form $f\left(x\right)=a\tan\left(bx-c\right)+d$`f`(`x`)=`a``t``a``n`(`b``x`−`c`)+`d` from a graph you will need to determine the values of the period first (this gives you $b$`b`) and then use that to find $c$`c` (using the phase shift). A small amount of algebraic manipulation will get help you find them.

How has the graph $y=5\tan x$`y`=5`t``a``n``x` been transformed from $y=\tan x$`y`=`t``a``n``x`?

Choose one of the following options:

Vertical dilation by a scale factor of $5$5.

AHorizontal dilation by a scale factor of $5$5.

BVertical dilation by a scale factor of $\frac{1}{5}$15.

CHorizontal translation by $5$5 to the right.

DVertical dilation by a scale factor of $5$5.

AHorizontal dilation by a scale factor of $5$5.

BVertical dilation by a scale factor of $\frac{1}{5}$15.

CHorizontal translation by $5$5 to the right.

D

Select all functions that have the same graph as $y=-\tan x$`y`=−`t``a``n``x`.

$y=-\tan\left(x+\frac{3\pi}{4}\right)$

`y`=−`t``a``n`(`x`+3π4)A$y=-\tan\left(x+\pi\right)$

`y`=−`t``a``n`(`x`+π)B$y=-\tan\left(x+\frac{\pi}{2}\right)$

`y`=−`t``a``n`(`x`+π2)C$y=-\tan\left(x+2\pi\right)$

`y`=−`t``a``n`(`x`+2π)D$y=-\tan\left(x+\frac{3\pi}{4}\right)$

`y`=−`t``a``n`(`x`+3π4)A$y=-\tan\left(x+\pi\right)$

`y`=−`t``a``n`(`x`+π)B$y=-\tan\left(x+\frac{\pi}{2}\right)$

`y`=−`t``a``n`(`x`+π2)C$y=-\tan\left(x+2\pi\right)$

`y`=−`t``a``n`(`x`+2π)D

The graph $y=\tan x$`y`=`t``a``n``x` (shown below) is reflected over the $y$`y`-axis and then translated vertically $3$3 units down.

Loading Graph...

Write the equation of the new graph.

Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions