# 2.2 Graphs of the other trigonometric functions  (Page 3/9)

 Page 3 / 9
$f\left(x\right)=A\mathrm{tan}\left(Bx-C\right)+D$

The graph of a transformed tangent function is different from the basic tangent function $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in several ways:

## Features of the graph of y = A Tan( Bx − C )+ D

• The stretching factor is $\text{\hspace{0.17em}}|A|.$
• The period is $\text{\hspace{0.17em}}\frac{\pi }{|B|}.$
• The domain is $\text{\hspace{0.17em}}x\ne \frac{C}{B}+\frac{\pi }{|B|}k,$ where $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is an integer.
• The range is $\text{\hspace{0.17em}}\left(\mathrm{-\infty },-|A|\right]\cup \left[|A|,\infty \right).$
• The vertical asymptotes occur at $\text{\hspace{0.17em}}x=\frac{C}{B}+\frac{\pi }{2|B|}k,$ where $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is an odd integer.
• There is no amplitude.
• $y=A\text{\hspace{0.17em}}\mathrm{tan}\left(Bx\right)\text{\hspace{0.17em}}$ is and odd function because it is the qoutient of odd and even functions(sin and cosine perspectively).

Given the function $\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bx-C\right)+D,\text{\hspace{0.17em}}$ sketch the graph of one period.

1. Express the function given in the form $\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bx-C\right)+D.$
2. Identify the stretching/compressing factor , $\text{\hspace{0.17em}}|A|.$
3. Identify $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ and determine the period, $\text{\hspace{0.17em}}P=\frac{\pi }{|B|}.$
4. Identify $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ and determine the phase shift, $\text{\hspace{0.17em}}\frac{C}{B}.$
5. Draw the graph of $\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bx\right)\text{\hspace{0.17em}}$ shifted to the right by $\text{\hspace{0.17em}}\frac{C}{B}\text{\hspace{0.17em}}$ and up by $\text{\hspace{0.17em}}D.$
6. Sketch the vertical asymptotes, which occur at where is an odd integer.
7. Plot any three reference points and draw the graph through these points.

## Graphing one period of a shifted tangent function

Graph one period of the function $\text{\hspace{0.17em}}y=-2\mathrm{tan}\left(\pi x+\pi \right)\text{\hspace{0.17em}}-1.$

• Step 1. The function is already written in the form $\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bx-C\right)+D.$
• Step 2. $\text{\hspace{0.17em}}A=-2,\text{\hspace{0.17em}}$ so the stretching factor is $\text{\hspace{0.17em}}|A|=2.$
• Step 3. $\text{\hspace{0.17em}}B=\pi ,\text{\hspace{0.17em}}$ so the period is $\text{\hspace{0.17em}}P=\frac{\pi }{|B|}=\frac{\pi }{\pi }=1.$
• Step 4. $\text{\hspace{0.17em}}C=-\pi ,\text{\hspace{0.17em}}$ so the phase shift is $\text{\hspace{0.17em}}\frac{C}{B}=\frac{-\pi }{\pi }=-1.$
• Step 5-7. The asymptotes are at $\text{\hspace{0.17em}}x=-\frac{3}{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=-\frac{1}{2}\text{\hspace{0.17em}}$ and the three recommended reference points are $\text{\hspace{0.17em}}\left(-1.25,1\right),\text{\hspace{0.17em}}$ $\left(-1,-1\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-0.75,-3\right).\text{\hspace{0.17em}}$ The graph is shown in [link] .

How would the graph in [link] look different if we made $\text{\hspace{0.17em}}A=2\text{\hspace{0.17em}}$ instead of $\text{\hspace{0.17em}}-2?$

It would be reflected across the line $\text{\hspace{0.17em}}y=-1,\text{\hspace{0.17em}}$ becoming an increasing function.

Given the graph of a tangent function, identify horizontal and vertical stretches.

1. Find the period $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ from the spacing between successive vertical asymptotes or x -intercepts.
2. Write $\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{tan}\left(\frac{\pi }{P}x\right).$
3. Determine a convenient point $\text{\hspace{0.17em}}\left(x,f\left(x\right)\right)\text{\hspace{0.17em}}$ on the given graph and use it to determine $\text{\hspace{0.17em}}A.$

## Identifying the graph of a stretched tangent

Find a formula for the function graphed in [link] .

The graph has the shape of a tangent function.

• Step 1. One cycle extends from –4 to 4, so the period is $\text{\hspace{0.17em}}P=8.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}P=\frac{\pi }{|B|},\text{\hspace{0.17em}}$ we have $\text{\hspace{0.17em}}B=\frac{\pi }{P}=\frac{\pi }{8}.$
• Step 2. The equation must have the form $f\left(x\right)=A\mathrm{tan}\left(\frac{\pi }{8}x\right).$
• Step 3. To find the vertical stretch $\text{\hspace{0.17em}}A,$ we can use the point $\text{\hspace{0.17em}}\left(2,2\right).$
$2=A\mathrm{tan}\left(\frac{\pi }{8}\cdot 2\right)=A\mathrm{tan}\left(\frac{\pi }{4}\right)$

Because $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{\pi }{4}\right)=1,\text{\hspace{0.17em}}$ $A=2.$

This function would have a formula $\text{\hspace{0.17em}}f\left(x\right)=2\mathrm{tan}\left(\frac{\pi }{8}x\right).$

Find a formula for the function in [link] .

$g\left(x\right)=4\mathrm{tan}\left(2x\right)$

## Analyzing the graphs of y = sec x And y = csc x

The secant    was defined by the reciprocal identity $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x=\frac{1}{\mathrm{cos}\text{\hspace{0.17em}}x}.\text{\hspace{0.17em}}$ Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at $\text{\hspace{0.17em}}\frac{\pi }{2},\text{\hspace{0.17em}}$ $\frac{3\pi }{2},\text{\hspace{0.17em}}$ etc. Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value.

We can graph $\text{\hspace{0.17em}}y=\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ by observing the graph of the cosine function because these two functions are reciprocals of one another. See [link] . The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.

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