document.write( "Question 695318: In my class we are graphing tangent functions and I missed the past two days.\r
\n" ); document.write( "\n" ); document.write( "In the equation... y=1/2tan2(x-5π/6)+1\r
\n" ); document.write( "\n" ); document.write( "I know the period= π/2
\n" ); document.write( "Horizontal asymptote= y=1
\n" ); document.write( "Steepness= 1/2
\n" ); document.write( "Phase shift = -5π/6\r
\n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #428767 by lwsshak3(11628)\"\" \"About 
You can put this solution on YOUR website!
y=1/2tan2(x-5π/6)+1
\n" ); document.write( "Equation for tan function:
\n" ); document.write( "y=tan(Bx-C), period=π/B, phase shift=C/B
\n" ); document.write( "Rewrite given equation in this form:
\n" ); document.write( "y=1/2tan(2x-10π/6)+1
\n" ); document.write( "B=2
\n" ); document.write( "period=π/B=π/2
\n" ); document.write( "1/4 period=π/8=3π/24
\n" ); document.write( "phase shift=C/B=(10π/6)/2=5π/6 (to the right)
\n" ); document.write( "vertical asymptote at x=13π/12+2πk,k=any integer (function undefined at this point)
\n" ); document.write( "horizontal asymptotes: none
\n" ); document.write( " \n" ); document.write( "
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