#1**+8 **

$$y=2+3sin\left(\frac{1}{2}(x-\pi)\right)$$

okay let's pull this apart.

The most important thing is that you have to know what y=sin(x) looks like! It has a wave length of 2pi

$$y=sin(ax) \mbox{ has a wavelength of } \frac{2\pi}{a}\quad

\mbox{ so }\\\\ y=sin\left(\frac{1}{2}x\right)\mbox{ has a wavelength of } \dfrac{2\pi}{\frac{1}{2}}=4\pi$$

$$\mbox{Now, the }\pi \mbox{ indicates a phase shift }\pi \mbox{ units to the positive side.}$$

$$\mbox{ Think of it like this } x-\pi=0 \mbox{ when } x=+\pi$$

so this is what we have so far.

$$y=2+3sin\left(\frac{1}{2}(x-\pi)\right)$$

okay, now we've done the hard bits.

$$y=1sin(x)\qquad \mbox{has a range of }\qquad -1\le y \le +1$$

$$y=3sin(x)\quad \mbox{ has a range of } -3\le y \le +3 \quad \mbox{ So that bit is easy! }$$

$$\mbox{ Now add 2 to all these y values and you 'lift' the curve so now the range will be }\quad -1\le y \le +5$$

NOW I will show you what this all looks like.

Melody Jun 2, 2014

#1**+8 **

Best Answer

$$y=2+3sin\left(\frac{1}{2}(x-\pi)\right)$$

okay let's pull this apart.

The most important thing is that you have to know what y=sin(x) looks like! It has a wave length of 2pi

$$y=sin(ax) \mbox{ has a wavelength of } \frac{2\pi}{a}\quad

\mbox{ so }\\\\ y=sin\left(\frac{1}{2}x\right)\mbox{ has a wavelength of } \dfrac{2\pi}{\frac{1}{2}}=4\pi$$

$$\mbox{Now, the }\pi \mbox{ indicates a phase shift }\pi \mbox{ units to the positive side.}$$

$$\mbox{ Think of it like this } x-\pi=0 \mbox{ when } x=+\pi$$

so this is what we have so far.

$$y=2+3sin\left(\frac{1}{2}(x-\pi)\right)$$

okay, now we've done the hard bits.

$$y=1sin(x)\qquad \mbox{has a range of }\qquad -1\le y \le +1$$

$$y=3sin(x)\quad \mbox{ has a range of } -3\le y \le +3 \quad \mbox{ So that bit is easy! }$$

$$\mbox{ Now add 2 to all these y values and you 'lift' the curve so now the range will be }\quad -1\le y \le +5$$

NOW I will show you what this all looks like.

Melody Jun 2, 2014