Thread: 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?

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?

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

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

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...

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 $\left(\dfrac{3}{2},2\right),\left(\dfrac{15}{2},2 \right)$.

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 $b = \dfrac{10\pi}{29}$).