Horizontal translation of y = 2cos(−½x + 90) − 3

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[tex]y=2\cos (\frac {-1} {2}x + 90 ) -3[/tex]

I changed the original function to

[tex]y=2\cos [\frac {-1} {2} (x-180)] -3[/tex]

period=-4pi
amplitude=2
max=-1
min=-5
range={y:-5<=y<=-1}
domain={x:XER}
vertical displacement 3 units down
horizontal translation 180 units left
 
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I don't think I am confused about radians and degrees some of my trig functions have radians in them so to make things simpler I convert to degrees but in this question i haven't done any converting that 90 is in degrees. :smile:
sorry I added the x in the original function.
 
Yes, seems correct although I think the horizontal displacement is to the right.
 
If the original question had 90 in it, then you did it correctly except for the direction of the horizontal translation. If the original question had [itex]\frac{\pi}{2}[/itex] in it, then the value of the horizontal translation is off too.
 
thanks so much the original question does have 90 degrees in it. You are right the horizontal translation should be to the right NOT LEFT! :smile: