Urgent! Finding function form Y= A sin(kx)+c

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Ramulala

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Hi there! Need urgent help on his graph. I have tried many times but couldn't able to get the my equation match the graph. I am somewhat behind on the learning content. It would be helpful if someone lets me know the answer.
Only 2 tries left to submit this question.
 

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After 11 posts you should know by now that you have to show your work in order to receive from tutors on this forum. You clearly have some work to show as you tried many times. Even if the work is wrong we get to see what method you want to use and where you are making any mistakes. Please follow the forum's posting guidelines and post back showing your work.
 
I'm really sorry. I was in a hurry.
In this question I know that it is a sin graph.

I know that amplitude is 2 and -4
I don't if this is how to write it 2/-4 or you have to do anything else to it??

Then I know that the complete graph starts at -12 and goes to positive 2. If I am not wrong you have to simplify it by 2π= 14. Which would give me π\7

so I think the final answer should be
y= 2/-4 sin( π\7x)

this is the best that I can explain right now. See if you can help me find the right answer. I don't think this is right.
 
Ramulala said:
I'm really sorry. I was in a hurry.
In this question I know that it is a sin graph.

I know that amplitude is 2 and -4
I don't if this is how to write it 2/-4 or you have to do anything else to it??

Then I know that the complete graph starts at -12 and goes to positive 2. If I am not wrong you have to simplify it by 2π= 14. Which would give me π\7

so I think the final answer should be
y= 2/-4 sin( π\7x)

this is the best that I can explain right now. See if you can help me find the right answer. I don't think this is right.
Click to expand...
 
You say In this question I know that it is a sin graph. That is not a good statement to make because there is no phase shift in the given equations. The sine graph you mention does not start at x=0!! The cosine graph does!

The amplitude is a real number, it is not two numbers.
The amplitude is (1/2)* (highest point - lowest point). Do you know which of A, C or k is the amplitude? If it is one of these variables do you really think that it can be two different values like 2 and -4?

One period of the given cosine graph starts at x=0 and finishes at x= ?? This causes k to equal??
 
So does that mean that amplitude will be 2. I am still not able to solve for k value?? Would it be found by finding the full graph and then simplifying it by by 2π\k. Or I am doing this part wrong.

There is not much time left to submit the answers. I am far behind the learning process in this. I need to watch and take lecture videos to understand it. If you know please provide more hints!!!
 
Ramulala said:
So does that mean that amplitude will be 2. I am still not able to solve for k value?? Would it be found by finding the full graph and then simplifying it by by 2π\k. Or I am doing this part wrong.

There is not much time left to submit the answers. I am far behind the learning process in this. I need to watch and take lecture videos to understand it. If you know please provide more hints or an answer !!!
Click to expand...
 
I am puzzled by this. Do you not have a text book that defines the "amplitude" of such a function? The amplitude is half the difference between the maximum value and the minimum value. If the function were the basic "A sin(x)" since sine has a maximum of 1 and a minimum of -1, Asin(x) has a maximum of A and a minimum of -A. A-(-A)= A+ A=2A so the amplitude is 2A/2= A. Here the maximum value is 2 and the minimum is -4. 2- (-4)= 2+ 4= 6 so the amplitude is 6/2= 3.
 
Ya! I was in a hurry and don't know the previous learning to find the answer, but after watching couple of videos I got it.

Answer is y= -3cos (π\5x - 5π/5) -1
Thanks to everyone. Who helped me. And thanks to jomo.
 
At the start, where x=0, you want kx=0 which is true for all k.

At the finish, where x=10, you want kx=2pi. That is k*10 = 2pi or k = 2pi/10 = pi/5. So you now have y = A(cos(pi/5*x))+C. A=3 (not -3, since A equaling a negative number implies that (in this case) the cosine graph you have is in fact a -cosine graph. You have the graph of a positive cosine.

How you got (π/5x - 5π/5) puzzles me since even if it is correct the problem demands that the angle be in the form of kx, NOT kx+b. So that angle has to be wrong as it is not in the correct format. Do you see that? Even if you do not know how to do this problem you need to see that your answer must be wrong. There is no question about it. It is in the wrong format!!
 
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Yes! I sense that very well, but since I was very late. I went on to Desmos software and tried different equations on it. It very well be that this random equation that put, matches the exact graph that was on the test.

I seriously need to go back to lecture notes and see the things that I am missing. Thankfully that equation worked and I am still shocked how I pulled it. Without knowing the whole concept.
 
Ramulala said:
Yes! I sense that very well, but since I was very late. I went on to Desmos software and tried different equations on it. It very well be that this random equation that put, matches the exact graph that was on the test.

I seriously need to go back to lecture notes and see the things that I am missing. Thankfully that equation worked and I am still shocked how I pulled it. Without knowing the whole concept.
Click to expand...
The equation you put is wrong-at least according to the instructions. They asked you to write the equation in the form y = Asin(kx) + C or y=A cos(kx)+C. You still don't see that your answer of y= -3cos (π\5x - 5π/5) -1 does not fit that form? OK, A=-3 and C = -1. Now what does k equal? Your angle must be in the form of kx. Your angle has a subtraction in it! Also you really should know that 5π/5=π. So π/5x - 5π/5 becomes π/5x - π
 
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