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

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Answer 1 · 2017-04-05T13:48:43

Please see below.

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]}

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