Need help solving this cosine equation from a graph?! Watch

Adorable98
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I have no idea how to solve parts a and b?
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1420787
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a) What do values P and Q mean for a trig function?

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Q gives the amplitude. This, being half of the distance from the too to the bottom of the function, is 4 here. (5-(-3))/2.


P is how much the function has been moved up or down. Here the trig function without P would go from 4 to -4, and has therefore been move up 1. So P is 1.



For b) Set the function equal to 0 and solve. This we do whenever we want to find roots.
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Adorable98
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(Original post by offhegoes)
a) What do values P and Q mean for a trig function?
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Q gives the amplitude. This, being half of the distance from the too to the bottom of the function, is 4 here. (5-(-3))/2.


P is how much the function has been moved up or down. Here the trig function without P would go from 4 to -4, and has therefore been move up 1. So P is 1.


For b) Set the function equal to 0 and solve. This we do whenever we want to find roots.
For Q why do we find the midpoint?
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1420787
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(Original post by Adorable98)
For Q why do we find the midpoint?
It works out the same, so the midpoint of the maximum value of 5 and the minimum value of -3 is 1, right?

Just a different way if doing it.
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Adorable98
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(Original post by offhegoes)
It works out the same, so the midpoint of the maximum value of 5 and the minimum value of -3 is 1, right?

Just a different way if doing it.
No, i mean't why do we find the midpoint?
And I also don't understand how you found P?
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(Original post by Adorable98)
No, i mean't why do we find the midpoint?
And I also don't understand how you found P?
Sorry, let me correct myself from before. For Q we do not find the midpoint. For Q we need to know that Q, because it multiplies the trig function, gives the amplitude, which is equal to half the distance from the top to the bottom. The distance from top to bottom is 8, so the amplitude is 4, which means Q is 4.

P, added to the trig function as it is, moves the function up (or down if negative). A normal trig function with nothing added will oscillate up and down about the x-axis, and the middle point will be 0. Here the middle point (midpoint between top and bottom) is 1, so the function has been moved up 1. So P is 1.
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Adorable98
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(Original post by offhegoes)
Sorry, let me correct myself from before. For Q we do not find the midpoint. For Q we need to know that Q, because it multiplies the trig function, gives the amplitude, which is equal to half the distance from the top to the bottom. The distance from top to bottom is 8, so the amplitude is 4, which means Q is 4.

P, added to the trig function as it is, moves the function up (or down if negative). A normal trig function with nothing added will oscillate up and down about the x-axis, and the middle point will be 0. Here the middle point (midpoint between top and bottom) is 1, so the function has been moved up 1. So P is 1.
Thaank you for explaining, but I seriously for some reason don't seem to understand it.
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Adorable98
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notnek Could you please help me with this question
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Adorable98
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Zacken Could you please explain this for me?
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felixastejada
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just thinking out of the box.
doesn't the equation look similar to y = mx + c ?
so since you have the y intercept -3, would that mean c = P = -3 ?
And then just substitute the values for y and x into the equation.
bring everything to one side by using inverse of cos and so on to get Q on its own?

ps. I'm probably very very wrong, just makes sense to me.
please somebody correct me
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username1732491
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(Original post by felixastejada)
just thinking out of the box.
doesn't the equation look similar to y = mx + c ?
so since you have the y intercept -3, would that mean c = P = -3 ?
And then just substitute the values for y and x into the equation.
bring everything to one side by using inverse of cos and so on to get Q on its own?

ps. I'm probably very very wrong, just makes sense to me.
please somebody correct me
It's not a linear graph, so you can't use y = mx + c
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felixastejada
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(Original post by TimGB)
It's not a linear graph, so you can't use y = mx + c
thought so
thanks
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username1732491
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(Original post by Adorable98)
Thaank you for explaining, but I seriously for some reason don't seem to understand it.
It's a lot simpler than it seems. You know the graph of cos(x) oscillates between -1 and 1, as does cos(2x) and cos(3x) and so on. The graph in question oscillates between -3 and 5, so you can use this to work out how much the graph has been stretched by (the value of Q) and how much it has been shifted by (the value of P).
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Adorable98
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(Original post by TimGB)
It's a lot simpler than it seems. You know the graph of cos(x) oscillates between -1 and 1, as does cos(2x) and cos(3x) and so on. The graph in question oscillates between -3 and 5, so you can use this to work out how much the graph has been stretched by (the value of Q) and how much it has been shifted by (the value of P).
I finally understood part (a)
but it's part (b) that I still don't understand
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username1732491
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(Original post by Adorable98)
I finally understood part (a)
but it's part (b) that I still don't understand
Probably should have also mentioned that you should check whether Q is negative by seeing if the graph is upside down or not - if it is, then Q must be negative.

For part b), set y = 0 and solve for x. You're looking for the first 6 positive x values.
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