Trigonometric Graphs

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

Transformations of cot, sec and cosec curves and equations

Lesson

We have seen before that different manipulations of a trigonometric function correspond to transformations of its graph. This is also true for the graphs of $y=\sec x$`y`=`s``e``c``x`, $y=\csc x$`y`=`c``s``c``x` and $y=\cot x$`y`=`c``o``t``x`.

In general, we can think about some function $y=f\left(x\right)$`y`=`f`(`x`). Remember that we can obtain the graph of $y=af\left(b\left(x-c\right)\right)+d$`y`=`a``f`(`b`(`x`−`c`))+`d`, where $a,b,c,d$`a`,`b`,`c`,`d` are constants, by applying a series of transformations to the graph of $y=f\left(x\right)$`y`=`f`(`x`). These transformations are summarised below.

Summary

To obtain the graph of $y=af\left(b\left(x-c\right)\right)+d$`y`=`a``f`(`b`(`x`−`c`))+`d` from the graph of $y=f\left(x\right)$`y`=`f`(`x`):

- $a$
`a`vertically dilates the graph of $y=f\left(x\right)$`y`=`f`(`x`) by a factor of $a$`a`. - $b$
`b`horizontally dilates the graph of $y=f\left(x\right)$`y`=`f`(`x`) by a factor of $\frac{1}{b}$1`b`. - $c$
`c`horizontally translates the graph of $y=f\left(x\right)$`y`=`f`(`x`) by $c$`c`units to the right. - $d$
`d`vertically translates the graph of $y=f\left(x\right)$`y`=`f`(`x`) by $d$`d`units upwards.

In the case that $a$`a` or $b$`b` is negative, there is also a reflection the graph of $y=f\left(x\right)$`y`=`f`(`x`) about the $x$`x`-axis or $y$`y`-axis, respectively.

In the case that $c$`c` or $d$`d` is negative, the direction of the translation is reversed.

When applying these facts to a cosecant, secant and cotangent, we need to consider how each type of transformation affects the key features of their graphs. In particular, it is helpful to consider features like vertical asymptotes, points of inflection, local minima and maxima, the median value of the function and the function's period.

If $f\left(x\right)$`f`(`x`) is $\operatorname{cosec}x$`c``o``s``e``c``x`, $\sec x$`s``e``c``x` or $\cot x$`c``o``t``x`, a horizontal dilation will change the period and a horizontal translation corresponds to a phase shift. A vertical translation will mean that all points on the graph are moved up or down by the same amount, while a vertical dilation will move all of the points on the graph away or towards the median value.

Plot the graph of $y=-2\sec x-1$`y`=−2`s``e``c``x`−1.

**Think:** We can think of the graph of this equation as a transformation of the graph of $y=\sec x$`y`=`s``e``c``x`. We can then perform the necessary transformations on the graph of $y=\sec x$`y`=`s``e``c``x` to arrive at the graph of $y=-2\sec x-1$`y`=−2`s``e``c``x`−1

**Do:** First, reflect the graph of $y=\sec x$`y`=`s``e``c``x` about the $x$`x`-axis. This is represented by applying a negative sign to the function (multiplying the function by $-1$−1).

Then we can dilate the function away from the $x$`x`-axis to match. This is represented by multiplying the $y$`y`-value of every point on $y=-\sec x$`y`=−`s``e``c``x` by $3$3.

Lastly, we translate the graph of $y=-2\sec x$`y`=−2`s``e``c``x` downwards by $1$1 unit, to obtain the final graph of $y=-2\sec x-1$`y`=−2`s``e``c``x`−1.

Write the equation of the graph of $y=\cot x$`y`=`c``o``t``x` after it has been horizontally dilated by a factor of $\frac{1}{3}$13 and then horizontally translated by $\frac{\pi}{4}$π4 units to the right.

**Think:** Below is a graph of the two transformations of the graphs. We want to manipulate $y=\cot x$`y`=`c``o``t``x` in the correct order to match up with the transformations.

**Do:** To horizontally dilate a function by a factor of $a$`a` we replace $x$`x` by $\frac{x}{a}$`x``a`. So the function will become $y=\cot3x$`y`=`c``o``t`3`x`.

To horizontally translate a function replace $x$`x` with $\left(x-b\right)$(`x`−`b`). So the function will then be in the form $y=\cot3\left(x-\frac{\pi}{4}\right)$`y`=`c``o``t`3(`x`−π4).

Careful!

The function $y=\cot\left(3x-\frac{\pi}{4}\right)$`y`=`c``o``t`(3`x`−π4) represents a horizontal shift by $\frac{\pi}{4}$π4 units and then a horizontal dilation by a factor of $\frac{1}{3}$13. The order matters as this would result in a different graph.

The graph $y=\sec x$`y`=`s``e``c``x` shown below is reflected over the $y$`y`-axis and then translated vertically $4$4 units up.

Loading Graph...

Write the equation of the new graph.

Plot the graph after the transformation has been applied.

Loading Graph...

Consider the function $y=\cot\left(3\left(x-\frac{\pi}{6}\right)\right)$`y`=`c``o``t`(3(`x`−π6)).

To get the graph of $y=\cot\left(3\left(x-\frac{\pi}{6}\right)\right)$

`y`=`c``o``t`(3(`x`−π6)) from $y=\cot x$`y`=`c``o``t``x`, we apply two transformations. Which transformation occurs first?Horizontal translation of $\frac{\pi}{6}$π6 units to the right.

AHorizontal dilation by a scale factor of $\frac{1}{3}$13.

BHorizontal translation of $\frac{\pi}{6}$π6 units to the right.

AHorizontal dilation by a scale factor of $\frac{1}{3}$13.

BThe graph of $y=\cot x$

`y`=`c``o``t``x`is given below. Adjust the points given to plot the graph of $y=\cot3x$`y`=`c``o``t`3`x`.Loading Graph...Your answer to part (b) is shown below. Adjust the points given to plot the graph of $y=\cot\left(3\left(x-\frac{\pi}{6}\right)\right)$

`y`=`c``o``t`(3(`x`−π6)).Loading Graph...

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