Graphing tangent curves with transformations

You can review the key features of tangent curves here.

Tangent Graphs

Recall that our standard tangent curve $y=\tan x$y=tanx looks like this.

 

It has the following key features:

• Vertical asymptotes at $\frac{\pi}{2}$π2​, $\frac{-\pi}{2}$−π2​, $\frac{3\pi}{2}$3π2​, $\frac{-3\pi}{2}$−3π2​, ...
• A period of $\pi$π between asymptotes
• A repeated shape between periods
• Multiple x-intercepts
• One y-intercept (at most)
• An infinite range
• A domain that excludes the x-values at the asymptotes

Applying certain constants to the equation to get $y=a\tan b\left(x-c\right)+d$y=atanb(x−c)+d will transform the graph $y=\tan x$y=tanx in certain ways.

Use the geogebra applet below to adjust the constants in $y=a\tan b\left(x-c\right)+d$y=atanb(x−c)+d and observe how it affects the graph. Try to answer the following questions.

• Which constants affect the position of the vertical asymptotes? Which ones don't?
• Which constants translate the graph, leaving the shape unchanged? Which ones affect the size?
• Which constants change the period of the graph? Which ones don't?
• Do any of these constants affect the range of the graph? If so, which ones?

The constants $a$a, $b$b, $c$c and $d$d transform the tangent graph in the following ways. You can review these transformations in more detail here.

• $a$a vertically dilates the graph.
  If $\left|a\right|<1$|a|<1 it will vertically compress the graph by a factor of $a$a and if $\left|a\right|>1$|a|>1 it will vertically stretch it by a factor of $a$a. If $a$a is negative, the graph will flip and reflect through the $x$x axis.
• $b$b horizontally dilates the graph, and therefore dilates the period.
  Unlike $a$a though, if $\left|b\right|<1$|b|<1 it will horizontally stretch the period by a factor of $\frac{1}{b}$1b​ and if $\left|b\right|>1$|b|>1 it will horizontally compress it by a factor of $\frac{1}{b}$1b​. If $b$b is negative, the graph will reflect through the $y$y axis.
• $c$chorizontally translates the graph $c$c units to the right, and is often referred to as the phase shift. If $c$c is negative, it translates the graph to the left.
• $d$d vertically translates the graph $d$d units upwards. If $d$d is negative, it translates the graph downwards.

 

Sketching tips

We can graph a tangent function by graphing the base function $y=\tan x$y=tanx and then applying the transformations in stages to achieve the final sketch. Just as with sine and cosine functions we could also sketch the function using key features. The following steps may assist in sketching:

1. Ensure you formula is in the required format, $y=a\tan\left(b\left(x-c\right)\right)+d$y=atan(b(x−c))+d rearrange if necessary. Write down $a$a, $b$b, $c$c and $d$d.
2. Sketch in a dotted line for $y=d$y=d. This is the central line where the points of inflection will lie. If the graph has not been vertically translated this will be the $x$x-axis.
3. Plot the point $\left(c,d\right)$(c,d). This is where the point of inflection at the origin of $y=\tan x$y=tanx has been translated to.
4.  Find the period: $P=\frac{\pi}{b}$P=πb​ .
5. From the point $\left(c,d\right)$(c,d) draw dotted lines for the asymptotes half a period in both directions. That is at $x=c+\frac{P}{2}$x=c+P2​ and  $x=c-\frac{P}{2}$x=c−P2​. Then at integer multiples of the period, $P$P, in each direction from these asymptotes until you have the required domain.
6. To achieve a more accurate sketch and clearly show the dilation, plot the points $\left(c+\frac{P}{4},d+a\right)$(c+P4​,d+a)  and $\left(c-\frac{P}{4},d-a\right)$(c−P4​,d−a) 
7. Draw a smooth curve through your points and approaching the asymptotes.
8. Repeat this smooth curve for each period.

 

Example

Sketch the function $y=3\tan\left(2\left(x-\frac{\pi}{4}\right)\right)+1$y=3tan(2(x−π4​))+1 for the interval $-\pi\le x\le\pi$−π≤x≤π

Think: What transformations would take $y=\tan x$y=tanx to $y=3\tan\left(2\left(x-\frac{\pi}{4}\right)\right)+1$y=3tan(2(x−π4​))+1? 

• We would dilate the graph by a factor of 3 from the $x$x-axis
• We need to dilate the graph by a factor of $\frac{1}{2}$12​ horizontally. Hence, the period becomes $\frac{\pi}{2}$π2​.
• We need to translate the graph by $1$1 unit vertically
• We need to translate the graph by $\frac{\pi}{4}$π4​ units horizontally

Do: List the parameters $a=3$a=3, $b=2$b=2, $c=\frac{\pi}{4}$c=π4​ and $d=1$d=1. Sketch a dotted line for the central line $y=1$y=1 and plot the point $\left(c,d\right)=\left(\frac{\pi}{4},1\right)$(c,d)=(π4​,1)

Find the period: $period=\frac{\pi}{b}$period=πb​$=\frac{\pi}{2}$=π2​ and draw dotted lines for the asymptotes half a period in both directions from the point $\left(\frac{\pi}{4},1\right)$(π4​,1). Then repeat at multiples of the period from these lines.

From the point $\left(\frac{\pi}{4},1\right)$(π4​,1) plot a point by going forwards $\frac{period}{4}$period4​$=\frac{\pi}{8}$=π8​ and up $a$a units ($3$3 units). Mirror this by plotting a second point backwards $\frac{\pi}{8}$π8​ from $\left(\frac{\pi}{4},1\right)$(π4​,1) and down $3$3 units.

Join the points with a smooth curve which also approaches the asymptotes. 

Lastly, repeat the pattern for each period.

Graph of $y=3\tan\left(2\left(x-\frac{\pi}{4}\right)\right)+1$y=3tan(2(x−π4​))+1

Reflect: Does the graph match how it should look after transformations? Does its cycle repeat the correct number of times for the domain given?

Axis intercepts

For our example above we did not have a $y$y-intercept as an asymptote coincided with the $y$y-axis. If this was not the case we could find the $y$y-intercept by evaluating the function at $x=0$x=0.

If we were required to label the $x$x-intercepts and there had been no vertical translation these would be found at the points of inflection at $x=c+Pn$x=c+Pn, where $P$P in the period and $n$n is any integer. However, if as in our example we have a vertical shift the $x$x-intercepts can be found with the assistance of technology or by solving the equation when $y=0$y=0. 

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3