Graph y=2tan(3x-30)+2 in degrees

Question

came across this problem and it is suppose to be graphed in degrees. I have only graphed in radians. How do I do it?

Answer 1

y = 2tan[3(x-10)] + 2

By transformation,

Graph one period of y1 = tan x , where x is in degrees. This is an odd function with vertical asymptotes at x = -90 and x = 90 degrees.

Stretch it vertically by a factor of 2 to get y2 = 2tan x

Compress it horizontally by a factor of 3 to get y3 = 2tan 3x

Move it to the right by 10 degrees to get y4 = 2tan [3(x-10)]

Move it up by 2 unites to get y5 = 2tan[3(x-10)] + 2 <==Answer

Answer 2

In radians the principal values for tangent go from -pi/2  to  pi/2.  In degrees this is from -90 to +90 degrees.  For this the argument of the tangent function must be in this domain.  Thus :

                                3x - 30 + 2  =  -90,   x=-62/3

                                3x - 30 + 2  =   90,   x=  118/3

                                  -62/3  ≤  x  ≤  118/3

To plot, plug in the values in this domain.  The range will run from -2∞  ≤  y  ≤  2∞  very similar to a radians plot.