In this discussion, we are going to graph the parent graph of the cotangent function,
discuss how to read the graph to come up with an equation,
and I will
explain how to use the standard equation to come up with the graph.
Here is an outline of this article:
1. Graph the Parent Graph of the Cotangent Function
o No Max and min, thus, no Amp
o Observe the Alternation of Asymptotes and Intercepts
o Find the Cycle given the Graph
o Find the Period given the Graph
o Find the Phase Shift Given the Graph
o Find the Vertical Shift Given the Graph
o Find the x-scale Given the Graph
2. The Standard Equation of the Cotangent Function
o Find the vertical Stretch/Compress Given the Standard Equation
o Find the Period and Horizontal Stretch/Compress Given the Standard Equation
o Find the Phase Shift Given the Standard Equation
o Find the Vertical Shift Given the Standard Equation
o Find the Cycle Given the Standard Equation
o Find the x-scale Given the Standard Equation
It is assumed that you already understand the unit circle.
Thus, we begin with what you already know.
Recall the unit circle:
We have terminal points in the form (x,y).
Cotangent takes the value x/y.
Thus,
Starting from zero and going counterclockwise,
cot(0) = 1/0 Undefined
cot(pi/4) = sqrt(2)/2/ sqrt(2)/2 = 1
cot(pi/2) = 0/1 = 0
cot(3pi/4) = -sqrt(2)/2/ sqrt(2)/2 = -1
cot(pi) = -1/0 Undefined
cot(5pi/4) = -sqrt(2)/2 / -sqrt(2)/2 = 1
cot(3pi/2) = 0/-1 = 0
cot(7pi/4) = sqrt(2)/2 / -sqrt(2)/2 = -1
cot(2pi) = 1/0 Undefined
We draw the graph of cotangent, connecting the dots and putting vertical asymptote everywhere cotangent is undefined.
Once you understand the shape of the cotangent graph, we only need to use the
quadrantal angles, which are the angles where its terminal points are along the x
and/or y axis.
The quadrantal angles on the unit circle are:
0, pi/2, pi, 3pi/2, and 2pi.
The pattern continues to the right and left.
Now, let us gather some information from the graph.
No Max, No Min, No Amplitude.
Since the range of the cotangent function is from negative infinity to positive
infinity, there are no max and min values (like sine and cosine has max and min
values). Thus, there is no amplitude.
Observing the Alternation between Intercepts and Asymptotes
Also, let us observe that the parent graph of the cotangent function alternates between intercepts and asymptotes.
The Cycle Given the Graph
Next, let us discuss the cycle of the cotangent graph.
The cycle of cotangent shows one complete period.
After the graph completes one cycle, it repeats itself for ever to the right and to the left.
From one asymptote to another, we have one cycle.
We see that the cycle begins at 0 and ends at pi.
Thus, the cycle is on the interval (0,pi).
Note that we have a set of parentheses.
This means that we are not actually including 0 or pi, but we can have values that
come very close to them.
The Period Given the Graph
Now, let us discuss the period.
The period of cotangent is the length of one cycle.
From 0 to pi, we have a length of pi.
Thus, period = pi
In our picture, we graphed the cotangent function over two periods.
The Phase Shift Given the graph
The phase shift, or the horizontal shift, tells us when the graph shifts to the right or
to the left.
Observing from left to right, we look for the location of the first asymptote to find
the phase-shift.
Do not confuse with the tangent graph. For tangent, the phase shift
begins at the intercept.
We see that our intercept begins at zero. Thus, the phase shift is zero.
The Vertical Shift Given the graph
The vertical shift of the graph tells us when the graph shifts up or down.
Observing up and down, we look for the location of the intercepts to find our vertical shift (it is not convenient to observe the asymptotes, as they are a straight line).
Since all intercepts are 0, we know that the vertical shift is zero.
The x-scale Given the Graph
The x-scale is the increment value along the x-axis.
Note that the critical points alternate between intercepts and asymptotes by a
length of pi/2.
Thus, the x-scale is pi/2.
Now, if you want to be more precise and find the midpoints between the intercepts and asymptotes, you can use an x-scale of pi4.
So, from the asymptote to the upper midpoint is a length of pi/4, from the upper midpoint to the intercept is a length of pi/4, and from the intercept to the lower midpoint is a le
ngth of pi/4, and from the lower midpoint the asymptote is a length of pi/4, and so on.
Many times, you will not need the midpoints. It is only useful to get an idea of the shape of the graph.
The Standard Equation of Cotangent
The standard equation of the cotangent function is of the form: y = acot[b(x-c)] + d.
If we were to write the original cotangent function in standard form, we have
y = acot[b(x-c)] + d.
y = 1cot[1(x-0)] + 0
Let us look at each part in detail.
The Vertical Stretch Given an Equation
|a| = 1: It is known at the vertical stretch/compress.
If |a| is greater than zero, but less than 1, the graph compresses vertically.
If |a| is greater than 1, the graph stretches vertically.
If “a” itself is less than zero, you will graph the function as if the negative sign did not exit, then reflect the x-axis at the end.
The Period and Horizontal Stretch/Compress Given an Equation
|b| = 1: It is known as the horizontal stretch/compress.
If |b| is greater than 0 but less than 1, the graph stretches horizontally.
If |b| is greater than 1, the graph compresses horizontally.
Notice that “b” operates in the opposite manner of “a.”
If “b” itself is less than zero, the graph reflects the y-axis of the original graph.
Or better yet, since cotangent is an odd function, we can simply pull out the negative sign and reflect the x-axis.
y = acot[-b(x-c)] + d = -acot[b(x-c)] + d
b is also related to the period by the following equation
Period = pi/|b| = pi/|1| =pi
The Phase Shift Given an Equation
c = 0: It is known as the phase shift, or horizontal shift.
This tells us how many units to shift right or left.
If c is greater than zero, we shift right.
If c is less than zero, we shift left.
Note, that in the form
y = acot[b(x-c)] + d , to find c,
set x-c=0 and solve for x to get x = c.
In the form
y = acot[b(x+c)] + d,
set (x+c)=0 and solve for x to get x = -c.
Notice that we flip the signs from the standard equation when finding c.
The Vertical Shift Given an Equation
d=0: It is known as the vertical shift.
This tells us how many units to shift up and down.
If d is greater than 0, we shift up
If d is less than zero, we shift down.
The main line of intercept is on the y = d.
We begin our graph at the phase shift. Thus, we begin at zero, which is an asymptote.
Don’t confuse the beginning of cotangent with tangent. With tangent, we begin at the intercept. With cotangent, we begin at the asymptote.
The Cycle Given an Equation
The cycle begins at “c ” and ends at “c + period”
Thus, we have cycle = (0, pi)
The x-scale Given an Equation
The x-scale is the increment values along the x-axis.
There are two ways to compute the x-scale. You can divide the period by 4 or you can divide the period by 2.
If you divide the period by 2, then you are calculating the increments of the x-scale alternating from the intercepts to the asymptotes.
If you divide the period by 4, then you are calculating the increments of the x-scale from the asymptote to the upper midpoint, and from the upper midpoint to the intercept, and from the intercept to the lower midpoint, and from the lower midpoint to the asymptote, and so on.
I prefer to divide by 2 and find the midpoints at the end (if necessary).
Many times, you may not need to compute the midpoints. So, it is easier to just find the intercepts and asymptotes and then compute the midpoints at the end if it is necessary. I will explain how as you keep reading.
To algebraically compute the x-scale of the original cotangent function, we say,
x-scale = period/2 = pi/2
Always begin your graph at “c.”
Our c is equal to 0.
Note that c is an asymptote.
Now add increments of the x-scale, which is pi/2, and follow the cotangent pattern.
The values are as follows:
0 (Asymptote)
0 + pi/2 = pi/2 (Intercept)
pi/2 + pi/2 = pi (Asymptote)
pi + 3pi/2 = 3pi/2 (Intecept)
3pi/2 + pi/2 = 2pi (Asymptote)
Draw your graph.
Now, if you want to find the upper and lower mid-points between the intercept and asymptote,
you can add the x-value of the intercept to the x-value of the asymptote (to the
right and to the left of the intercept), then divide by 2.
This will give you the placement of the x value.
So, I will add pi/2 + 0 = pi/2.
Then divide pi/2 by 2, which gives us pi/4.
We see that the midpoint lines up with pi/4.
Similarly, we add pi + pi/2 = 3pi/2.
Then divide by 2, which gives us 3pi/4.
We see that the midpoint lines up with 3pi/4.
The height of the upper midpoint will take the value of d+ |a| = 0 +1 =1
The height of the lower midpoint will take the value of d- |a| = 0 -1 = -1
It is important to understand the parent graph to know when to add or subtract |a|.
We see that the values above are on the parent graph of cotangent.
Now you are ready to graph the cotangent function with multiple transformations!
Thank you for reading! I hope that I can assist you on your math journey!
Sincerely,
MathAngel369
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