Just as we transformed the trigonometric functions $y=\sin x$y=sinx and $y=\cos x$y=cosx we can apply parameters to the equation $y=\tan x$y=tanx to transform it to $y=a\tan\left(b\left(x-c\right)\right)+d$y=atan(b(x−c))+d.
Use the geogebra applet below to adjust the parameters in $y=a\tan\left(b\left(x-c\right)\right)+d$y=atan(b(x−c))+d and observe how it affects the graph. Try to answer the following questions.
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The constants $a$a, $b$b, $c$c and $d$d transform the tangent graph. Let's summarise the impact of each:
To obtain the graph of $y=a\tan\left(b\left(x-c\right)\right)+d$y=atan(b(x−c))+d from the graph of $y=\tan\left(x\right)$y=tan(x):
Illustrate the change in dilation of the graph $y=\tan x$y=tanx by sketching the graphs of $y=\tan x$y=tanx together with $y=3\tan x$y=3tanx and $y=\frac{2}{7}\tan x$y=27tanx.
Think: Note since only a vertical dilation has been applied all three graphs will share $x$x-intercepts and vertical asymptotes. We first sketch the base graph of $y=\tan x$y=tanx, shown in blue in the graph below. This graph will go through the points $\left(\frac{\pi}{4},1\right)$(π4,1), $\left(0,0\right)$(0,0) and $\left(\frac{-\pi}{4},-1\right)$(−π4,−1). The asymptotes will be located at $\frac{\pi}{2}$π2, $\frac{-\pi}{2}$−π2, $\frac{3\pi}{2}$3π2, $\frac{-3\pi}{2}$−3π2,....
The graph of $y=3\tan x$y=3tanx has been vertically dilated by a factor of $3$3. So we can plot the points $\left(\frac{\pi}{4},3\right)$(π4,3) and $\left(\frac{-\pi}{4},-3\right)$(−π4,−3) to show this stretch clearly. Similarly the graph $y=\frac{2}{7}\tan x$y=27tanx has been vertically dilated by a factor of $\frac{2}{7}$27. To show this we can plot the points $\left(\frac{\pi}{4},\frac{2}{7}\right)$(π4,27) and $\left(\frac{-\pi}{4},\frac{-2}{7}\right)$(−π4,−27).
Do: The graphs together are shown below.
Sketch the graph of $f(x)=\tan\left(x-\frac{\pi}{4}\right)$f(x)=tan(x−π4).
Think: We see that the function $\tan x$tanx has been moved to the right by a distance of $\frac{\pi}{4}$π4 units. We can sketch this by drawing the base graph of $y=\tan\theta$y=tanθ and shifting each point right by $\frac{\pi}{4}$π4.
The phase shift will also move the asymptotes and since $\tan x$tanx is undefined at $x=\frac{\pi}{2}+n\pi$x=π2+nπ for all integers $n$n, the undefined points for $\tan\left(x-\frac{\pi}{4}\right)$tan(x−π4) must be $x=\frac{3\pi}{4}+n\pi$x=3π4+nπ.
Do: The graph is shown below in purple.
Consider the function $y=-4\tan\frac{1}{5}\left(x+\frac{\pi}{4}\right)$y=−4tan15(x+π4).
Determine the period of the function, giving your answer in radians.
Determine the phase shift of the function, giving your answer in radians.
Determine the range of the function.
$[-1,1]$[−1,1]
$(-\infty,0]$(−∞,0]
$[0,\infty)$[0,∞)
$(-\infty,\infty)$(−∞,∞)
The graph of $y=\tan x$y=tanx is shown below. On the same set of axes, draw the graph of $y=5\tan x$y=5tanx.
Select all functions that have the same graph as $y=-\tan x$y=−tanx.
$y=-\tan\left(x+\frac{3\pi}{4}\right)$y=−tan(x+3π4)
$y=-\tan\left(x+\pi\right)$y=−tan(x+π)
$y=-\tan\left(x+\frac{\pi}{2}\right)$y=−tan(x+π2)
$y=-\tan\left(x+2\pi\right)$y=−tan(x+2π)