1. Graph the trigonometric function...

When I enter my equation into my graphing calculator it's not looking like the graph in my attachment. I would like to know where I am going wrong..

So what I do know is that this graph is of the form y= asin(bx + c) +d. The graph has been raised by one unit so I know that d =1 and that a=1 as well. The period of the function is calculated by using the formula p= 2pi/b. So to figure this out I looked to see where this graph completed one full cycle. It looks like the cycle begins at x =-1 to x = 6. So I set my period to be 7 and from that I solved for b. b= 2pi/7. To find my phase shift I set bx+c=0 and from that c= 2pi/7? My final answer then for this problem is y = sin((2pi/7)× +2pi/7)+ 1? So where did i go wrong?

2. Re: Graph the trigonometric function...

When I graph $\displaystyle sin(2\pi x/7+ 2\pi/7)$ it looks very much like the graph you show. What are you getting? Is your calculator in radian mode?

3. Re: Graph the trigonometric function...

To me, it looks like $y=1$ when $x=0,x=6$. So, I think it would be

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

4. Re: Graph the trigonometric function...

Ok so the calculator I'm using is actually an app called mathway. So that could be the problem. When I went back to my answer I plugged in points that are on the graph and got different values. The points I used were(2,2) and (8,2) and they didnt check out. But maybe I can try to get an actual calculator and use degree mode when graphing this problem...

5. Re: Graph the trigonometric function...

Originally Posted by slapmaxwell1
Ok so the calculator I'm using is actually an app called mathway. So that could be the problem. When I went back to my answer I plugged in points that are on the graph and got different values. The points I used were(2,2) and (8,2) and they didnt check out. But maybe I can try to get an actual calculator and use degree mode when graphing this problem...
I would say the points would be $\left(\dfrac{3}{2},2\right),\left(\dfrac{15}{2},2 \right)$.

6. Re: Graph the trigonometric function...

Actually, it is easier to see where it hits zero than it is to see where it hits 2. It looks like it hits zero at $x=\dfrac{87}{20}$ and $x=\dfrac{111}{20}$. But, that will be a much uglier solution.

You wind up with

$$y=\sin \left(\dfrac{10\pi x}{29} \right) + 1$$

This is because at $x=0$, $y=1$ (so $c=0$) and $x=\dfrac{87}{20}$, $y = 0 = \sin\left(\dfrac{3\pi}{2}\right)+1$ (so $b = \dfrac{10\pi}{29}$).