Question 695318: In my class we are graphing tangent functions and I missed the past two days.
In the equation... y=1/2tan2(x-5π/6)+1
I know the period= π/2
Horizontal asymptote= y=1
Steepness= 1/2
Phase shift = -5π/6
Now for my asymptote I ended up with ananswer of 13π/12+π/2k,kei. I do not feel my math is correct and wanted to know if that was the true answer.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! y=1/2tan2(x-5π/6)+1
Equation for tan function:
y=tan(Bx-C), period=π/B, phase shift=C/B
Rewrite given equation in this form:
y=1/2tan(2x-10π/6)+1
B=2
period=π/B=π/2
1/4 period=π/8=3π/24
phase shift=C/B=(10π/6)/2=5π/6 (to the right)
vertical asymptote at x=13π/12+2πk,k=any integer (function undefined at this point)
horizontal asymptotes: none
|
|
|
