Question 446692: Graph and label y= 2 tan(3x-2π/3)+5
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Graph and label y= 2 tan(3x-2π/3)+5
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y= 2 tan(3x-2π/3)+5
Standard form for tan function: tan(Bx-C), Period =π/B, phase-shift=C/B
B=3
Period=π/3=12π/36
1/4 Period=3π/36
Phase-shift
set 3x-(2/3)π=0
3x=2π/3
x=2π/9=8π/36 (phase-shift) to right
You now have the information you need to graph the given function. I don't have the means to graph it for you but I will show you how to do it as follows: (For ease of calculation, I have used 36 as the common denominator. In the final answer you should reduce the fractions to lowest terms.)
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On the x-axis mark 8π/36. To the right make 2 more marks spaced 1/4 period apart, and to the left make 2 more marks spaced 1/4 period apart, that is, you will now have the following marks on the x-axis: 2π/36, 5π/36, 8π/36, 11π/36, 14π/36 Note the the difference between the first and last mark is 14π/36-2π/36=12π/36=one period.
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solving for y
at x=2π/36
y=2tan(3x-2π/3)+5
y=2tan(π/6-4π/6)+5
y=2tan(-π/2)+5=undefined (tan -π/2 is undefined)
at x=5π/36
y=2tan(5π/12-8π/12)+5
y=2tan(-π/4)+5=-2+5=3
at x=8π/36
y=2tan(8π/12-8π/12)+5
y=2tan(0)+5
y=5
at x=11π/36
y=2tan(11π/12-8π/12)+5
y=2tan(π/4)+5=2+5=7
at x=14π/36
y=2tan(14π/12-8π/12)+5
y=2tan(6π/12)+5
y=2tan(π/2)+5=undefined (tan π/2 is undefined)
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You now have the following points to draw your graph of the given tan function:
First, you have vertical asymptotes at x=2π/36 and x=14π/12.
Points: (2π/36,-∞), (5π/36,3), (8π/36,5), (11π/36,7), and (14π/36,∞)
This seems like a lot of algebra, but in effect all you are doing is multiplying the basic tan curve by 2, shifting it by 8π/36 to the right and bumping it up 5 units.
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