hey guys how to draw trigo graph, question below:
the function f is defined, for 0<= x<= pie radians
f(x)= 3 sin (x/2) -2
my thought process is the range of x now is 0<=x/2<= pie radian/2
so x/2 values are pie radian/8, pie radian/4, 3 pie radian/8 and pie radian/2
y values corresponding are -0.852, 0.121, 0.772 and 1
when i sketch the curve, it is a decreasing function, is it correct
Hi,
The usual sine curve has an amplitude of 1 unit and a period of 2 pi radians, so we have
0 <= x/2 <= 2 pi, which leads to 0 <= x <= 4 pi.
When we sketch y = f(x) for the interval [0, pi], it'll be a quarter of the original sine curve. Actually, we are stretching the curve horizontally so that it is twice as long as before.
Next, we stretch the sine curve vertically so that its amplitude is 3 units (i.e. thrice as tall as before).
Finally, we move the entire curve 2 units down.
Thanks!
Cheers,
Wen Shih