How do you graph # y=2+1/2sin2(x-5)#?

1 Answer
Apr 5, 2017

Answer:

Please see below.

Explanation:

As we have #y=2+1/2sin2(x-5)# and any sine ratio has maximum value of #+1# and minimum value of #-1#,

maximum value of #y# will be #2+1/2=2 1/2# and minimum value will be #2-1/2=1 1/2#. As such there is no #x#-intercept.

Hence #y# will move between these two numbers.

Maxima #2 1/2# is there when #2(x-5)=(4n+1)/2pi# i.e. at #x=(4n+1)/4pi+5=npi+pi/4+5# and some values are #x={-0.4978,2.6438,5.7854,8.927,12.0686}#

Minima #1 1/2# is there when #2(x-5)=(4n-1)/2pi# i.e. at #x=(4n-1)/4pi+5=npi-pi/4+5# and some values are #x={-2.0686,1.073,4.2146,7.3562,10.4978}#

Mean value #2# appears at #2(x-5)=npi# i.e. #x=n/2pi+5# and some values are #x={0.2876,1.8584,3.4292,5,6.5708,8.1416,9.7124}#

Now when #x=0#, we have #y=2+1/2xxsin(-5)# ad considering it in radiansm #y=2+1/2xx0.9589=2.4795# and hence #y# intercept is #~=2.48#

and function appears as follows

graph{2+(1/2)sin(2x-10) [-1.46, 8.54, -0.58, 4.42]}