came across this problem and it is suppose to be graphed in degrees. I have only graphed in radians. How do I do it?
Graph y=2tan(3x-30)+2 in degrees
y = 2tan[3(x-10)] + 2
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
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.