SOLUTION: Im working on trigonometric function and not sure how solve or graph these the problem is y= 3 cot(x+pie over 4). I have tried to understand but i am lost thank you

Question 428515: Im working on trigonometric function and not sure how solve or graph these the problem is y= 3 cot(x+pie over 4). I have tried to understand but i am lost thank you
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Im working on trigonometric function and not sure how solve or graph these the problem is
y= 3 cot(x+pie over 4)
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The first thing you need to know is the basic curve of the cot function, y=cotx. Let me try to describe it to you. Start with an (x,y) coordinate system with the x-axis marked for radians.
The cot function is defined as cosx/sinx. When x=0, sinx=0, therefore, the cot function is undefined at x=0. The y-axis becomes a vertical asymptote for cotx as it heads into infinity as x approaches zero. At x=pi/4,cotx=1. Further along at x=pi/2, cotx=0. Then at x=3pi/2, cotx=-1. As x approaches pi, sinx again approaches zero which makes cotx approaching negative infinity with x=pi radians as the asymptote. What I have described is one full period which is pi. This function repeats itself every pi radian.
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Now, for your problem:
The standard form for the cot function is Acot(Bx-C), with period=2pi/B, phase shift C/B, and A being a multiplier which would make the curve steeper the higher the multiplier.
Given equation: y= 3cot(x+pi/4)
B=1
period=2pi/1=2pi
1/4 period=2pi/4=pi/2
C=pi/4
phase shift =C/B=pi/4
This means you will be starting your curve at x=-pi/4 which is the vertical asymptote. 1/4 of a period to the right at x=-pi/4+2pi/4=pi/4, cotx =3 with the multiplier, 3.The next point to mark on the x-axis is further along at x=pi/4+2pi/4=3pi/4 where cotx=0. Next we add another 1/4 period, x=3pi/4+2pi/4=5pi/4, where cosx=-3 with the multiplier. Finally, we move another 1/4 period, x=5pi/4+2pi/4=7pi/4, where cosx approaches negative infinity. x=7pi/4 is an asymptote. Note that we plotted the curve from x=-pi/4 to x=7pi/4, which covers a total of 8pi/4 or 2pi, the period of the function. This curve of cotx should look exactly like the basic curve described above except it was shifted pi/4 to the left and became steeper with a multiplier of 3.
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I'm afraid without diagrams this explanation is probably not very clear, but I thought I would give it a shot.
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