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

Apr 5, 2017

Explanation:

As we have $y = 2 + \frac{1}{2} \sin 2 \left(x - 5\right)$ and any sine ratio has maximum value of $+ 1$ and minimum value of $- 1$,

maximum value of $y$ will be $2 + \frac{1}{2} = 2 \frac{1}{2}$ and minimum value will be $2 - \frac{1}{2} = 1 \frac{1}{2}$. As such there is no $x$-intercept.

Hence $y$ will move between these two numbers.

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

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

Mean value $2$ appears at $2 \left(x - 5\right) = n \pi$ i.e. $x = \frac{n}{2} \pi + 5$ and some values are $x = \left\{0.2876 , 1.8584 , 3.4292 , 5 , 6.5708 , 8.1416 , 9.7124\right\}$

Now when $x = 0$, we have $y = 2 + \frac{1}{2} \times \sin \left(- 5\right)$ ad considering it in radiansm $y = 2 + \frac{1}{2} \times 0.9589 = 2.4795$ and hence $y$ intercept is $\cong 2.48$

and function appears as follows

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