Question 1112426
<pre>
The other tutor's answer is correct about the period, but
incorrect about the vertical asymptotes.
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The period of 

{{{f(x)}}}{{{""=""}}}{{{a*tan(bx)}}}

is {{{pi/b}}}

and has vertical asymptotes wherever 
bx is an odd multiple of {{{pi/2}}}

Therefore the period of

{{{f(x)}}}{{{""=""}}}{{{3tan(2x)}}}

is {{{pi/2}}}

and has vertical asymptotes wherever 
2x is an odd multiple of {{{pi/2}}}

Odd numbers are represented by 2n-1, so 
between -<font face="symbol">p</font> and +<font face="symbol">p</font>,

{{{2x = (2n-1)expr(pi/2)}}}

{{{x=expr((2n-1)/4)pi

{{{-pi<=expr((2n-1)/4)pi<=pi}}}

Multiplying through by 4

{{{-4pi<=(2n-1)pi<=4pi}}}

Dividing all three sides by <font face="symbol">p</font>

{{{-4<=2n-1<=4}}}

Adding 1 to all three sides:

{{{-3<=2n<=5}}}

Dividing all three sides by 2

{{{-3/2<=n<=5/2}}}

And since n is an integer, n=-1,0,1,2

So when n=-1, (2n-1)pi/2 = -3<font face="symbol">p</font>/4
And when n=1, (2n-1)pi/2 = -<font face="symbol">p</font>/4
And when n=1, (2n-1)pi/2 = <font face="symbol">p</font>/4
And when n=1, (2n-1)pi/2 = 3<font face="symbol">p</font>/4

So in the interval (-<font face="symbol">p</font>,<font face="symbol">p</font>),
the graph continues through 4 periods, and there are 4 vertical 
asymptotes at: 

x = -3<font face="symbol">p</font>/4, approximately -2.4
x = -<font face="symbol">p</font>/4, approximately -0.8
x = <font face="symbol">p</font>/4, approximately 0.8
x = 3<font face="symbol">p</font>/4, approximately 2.4

the 4 green lines are the 4 vertical asymptotes:

{{{drawing(200,400,-3.5,3.5,-7,7,
graph(200,400,-3.5,3.5,-7,7,3tan(2x)), green(line(-3pi/4,-10,-3pi/4,10), line(-pi/4,-10,-pi/4,10),line(pi/4,-10,pi/4,10),line(3pi/4,-10,3pi/4,10)) )}}}

Edwin</pre>