TEF - Graphing Tangent Lesson

Graphing Tangent

Let's think about how to graph $f\left(t\right)=\tan\left(x\right)$.

We know that tangent represents a specific ratio:

Tangent: the ratio of y to x, $\tan\theta=\frac{y}{x}$

Let's use our Unit Circle values to create a table:

 

t

0

$\frac{\pi}{6}$

$\frac{\pi}{4}$

$\frac{\pi}{3}$

$\frac{\pi}{2}$

$\frac{2\pi}{3}$

$\frac{3\pi}{4}$

$\frac{5\pi}{6}$

$\pi$

$\frac{7\pi}{6}$

$\frac{5\pi}{4}$

$\frac{4\pi}{3}$

$\frac{3\pi}{2}$

$\frac{5\pi}{3}$

$\frac{7\pi}{4}$

$\frac{11\pi}{6}$

$2\pi$

$f\left(t\right)=\tan t$

0

$\frac{\sqrt[]{3}}{3}\approx0.577$

1

$\sqrt[]{3}\approx1.732$

undefined

$-\sqrt[]{3}\approx-1.732$

-1

$-\frac{\sqrt[]{3}}{3}\approx-0.577$

0

$\frac{\sqrt[]{3}}{3}\approx0.577$

1

$\sqrt[]{3}\approx1.732$

undefined

$-\sqrt[]{3}\approx-1.732$

-1

$-\frac{\sqrt[]{3}}{3}\approx-0.577$

0

 

 

 

 

 

Notice that in the table above, when the angle is $\frac{\pi}{2}\:or\:\frac{3\pi}{2},$ the output of tangent is undefined. In addition, there are no points on the graph at those values. That is because:

$\tan\left(\frac{\pi}{2}\right)=\frac{y}{x}=\frac{1}{0}=undefined\:and\:\tan\left(\frac{3\pi}{2}\right)=\frac{y}{x}=-\frac{1}{0}=undefined$

 

So, at those two values, there are vertical asymptotes.

 

 

Since the period above is , there are going to be asymptotes at $x=\frac{\pi}{2}$ and all values that are multiples of  away. This means there is an asymptote at all values of $x=\frac{\pi}{2}+\pi n,\:where\:n\:is\:an\:integer.$ Since the tangent graph does not contain maximum or minimum values, there is no amplitude. 

Let's identify some important features of $y=\tan x$

1. What is the domain of $y=\tan x$?

• Solution: all real numbers,  $x\ne\frac{\pi}{2}+\pi n$

2. What is the range of $y=\tan x$?

• Solution: all real numbers

3. What is the period of $y=\tan x$?

• Solution: $\pi$

Watch this video to work on graphing some transformations of $y=\tan x$.

 

Let's try graphing another tangent function: $f\left(x\right)=\tan\left(x+\frac{\pi}{2}\right)$.

First, let's identify the key features:

1. There is no vertical stretch because A = 1.

2. The period is $\pi$ because B = 1, so $\frac{\pi}{B}=\frac{\pi}{1}=\pi$.

3. To solve for the asymptotes, you set up the equation:

$\frac{\pi}{2}=Bx+C \\ \frac{\pi}{2}=1x+\frac{\pi}{2} \\ 0=x$

So, the first asymptote is at x = 0 and all the other asymptotes are at intervals of $\pi$ units away. So the graph of the function should look like this:

 

 

 

 IMAGES CREATED BY GAVS