Trigonometry: Analyzing and Graphing a Cosine Function

B = πC = 3π/2Horizontal Shift = -(3/2)B = πC = 3π/2Horizontal Shift = -(3/2)Yes, so your B and C values are correct. But your horizontal shift is incorrect. Can you tell me what the horizontal shift should be?The horizontal shift should be 0 because the graph starts at 0 and shifts to the left by 0.In summary, the problem involves finding the amplitude, period, horizontal shift, vertical translation, and step of a given equation and graphing it. Using the formula y = A cos (Bx + C) + D, the amplitude is 1/3, the period
  • #1
EddieLP
5
0
Hey I need a second opinion to see if my formulas and calculations are correct. Also I need help with graphing this problem.

My problem is to find amplitude, period, horizontal shift, vertical translation, step and then graph of:

y= 2 - (1/3) cos ( πx + (3π/2) ) , -1/4 ≤ x ≤ 15/4

Using the formula:

y = A cos ( Bx + C) + D

A = 2

B = π

C = 3/2

D = 0

Amplitude: |A| = 1/3
Period: ((2π) / |B|) = 2
Horizontal Shift: -C/B = (3/(2π))
Vertical Translation: D = 0
Step: ((2π)/2)/B = 1
 
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  • #2
All of your data results appear to be correct.
Mod edit: Some of the results are incorrect.
What is the problem you are having with the graph?
 
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  • #3
JBA said:
All of your data results appear to be correct. What is the problem you are having with the graph?
I'm lost on what goes where on the graph.
 
  • #4
Do you know the general shape of a simple y=cos x curve?

What method are you using to try to create this curve?

Actually, upon review I have realized that your A = 2 and D = 0 values are in error but your amplitude is correct. Look at all of the equation modifiers again to see if you can identify the correct values for A & D

Also review your C/B value with respect to π
 
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  • #5
EddieLP said:
Hey I need a second opinion to see if my formulas and calculations are correct. Also I need help with graphing this problem.

My problem is to find amplitude, period, horizontal shift, vertical translation, step and then graph of:

y= 2 - (1/3) cos ( πx + (3π/2) ) , -1/4 ≤ x ≤ 15/4

Using the formula:

y = A cos ( Bx + C) + D
A = 2
B = π
C = 3/2
D = 0
A is not 2 and D is not 0.
EddieLP said:
Amplitude: |A| = 1/3
Period: ((2π) / |B|) = 2
Horizontal Shift: -C/B = (3/(2π))
Vertical Translation: D = 0
Step: ((2π)/2)/B = 1
Your amplitude and period are correct, but the horizontal shift and vertical translation are incorrect. I don't know what "step" means with regard to this problem. Is "step" half a period?
 
  • #6
Amplitude: |A| = 1/3
Period: ((2π) / |B|) = 2
Horizontal Shift: -C/B = - (3/(2π))
Vertical Translation: D = 0
Step: ((2π)/2)/B = 1

I've made adjustments to Horizontal Shift by making it a negative number
I put Vertical translation as 0 since there is no D in the problem.
I'm not sure what step is but our teacher told us the formula for it.

Is there something I am missing to find Vertical translation?
Is there another name for "Step"?
The formulas that were given to me for step are different for each function. Sin and Cos is: ((2π)/2)/B, Tan is (π/2)/B
 
  • #7
OK, your problem is thinking that D must be at the end of the equation. Reverse the equation order and see what that makes the value of D be.
 
  • #8
EddieLP said:
Amplitude: |A| = 1/3
Period: ((2π) / |B|) = 2
Horizontal Shift: -C/B = - (3/(2π))
No, that's still wrong.
EddieLP said:
Vertical Translation: D = 0
Still wrong.
EddieLP said:
Step: ((2π)/2)/B = 1

I've made adjustments to Horizontal Shift by making it a negative number
What is B? What is C?
EddieLP said:
I put Vertical translation as 0 since there is no D in the problem.
Yes, there is.
EddieLP said:
I'm not sure what step is but our teacher told us the formula for it.

Is there something I am missing to find Vertical translation?
Is there another name for "Step"?
The formulas that were given to me for step are different for each function. Sin and Cos is: ((2π)/2)/B, Tan is (π/2)/B
This is because the period of the basic, untransformed sine and cosine functions is ##2\pi##. The period for the basic tangent function is ##\pi##.
 
  • #9
JBA said:
OK, your problem is thinking that D must be at the end of the equation. Reverse the equation order and see what that makes the value of D be.
I reversed the order and got 2!
So Vertical Translation = 2

Mark44 said:
No, that's still wrong.
Still wrong.
What is B? What is C?
I found B and C by factoring out p from:

y= 2 - (1/3) cos ( πx + (3π/2) )

factored form:

y= 2 - (1/3) cos ( π ( x + (3/2) )

from the formula: y = A cos ( Bx + C) + D
B = π
C = 3/2

Formula for Horizontal Shift is - ( C / B )
(3/2)/(π/1) = 3/(2π)

I'm trying to find out what's missing, still trying to find. any hints?
 
  • #10
EddieLP said:
I reversed the order and got 2!
So Vertical Translation = 2
Yes
EddieLP said:
I found B and C by factoring out p from:

y= 2 - (1/3) cos ( πx + (3π/2) )

factored form:

y= 2 - (1/3) cos ( π ( x + (3/2) )

from the formula: y = A cos ( Bx + C) + D
B = π
C = 3/2
No, the pattern in your formula is different.
Formula: y = A cos(Bx + C)
In your work, you factored out ##\pi## from both terms
The formula is NOT A cos(B(x + C))
EddieLP said:
Formula for Horizontal Shift is - ( C / B )
(3/2)/(π/1) = 3/(2π)
This formula, which you are blindly applying, is for the equation y = A cos (Bx + C)
From this formula, what is B and what is C?
EddieLP said:
I'm trying to find out what's missing, still trying to find. any hints?
 
  • #11
The vertical shift is 2; so now you have everything except a confirmed horizontal shift.

Actually, you have to think in terms of radians, the simple answer is that C = 3π/2 = 1.5π = 270° such that when x = 0 then cos (0 + 270°) = -1 and with the vertical y offset = 2 the result is 2 -1 = 1 (the minimum height of the oscillation) and when x = .5 then cos (.5 + 1.5)π = 2π = cos 180° =0 and y = 2+0 = 2 (the centerline of the curve oscillation) and when x = 1, cos ( 1 +1.5)π = cos 2.5 = cos 90° = 1 and y = 2+1 = 3 ( the maximum height of the oscillation), etc
 
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  • #12
One clarification on the above, all of the above max/ min values of 1 & -1 are before muliplying the results of the cos by 1/3 as required by the total equation so for the purposes of your calculations and graph the results are actually 2 - 1/3 x 1 = 1.666, 2 - 1/3 x 0 = 2 and 2 - 1/3 X -1 = 2 + 1/3 = 2.333.
 
  • #13
Mark44 said:
Yes
No, the pattern in your formula is different.
Formula: y = A cos(Bx + C)
In your work, you factored out ##\pi## from both terms
The formula is NOT A cos(B(x + C))
This formula, which you are blindly applying, is for the equation y = A cos (Bx + C)
From this formula, what is B and what is C?
B = πx
C = 3π/2
Horizontal Shift = -(3/2)!
My book says that I have to factor out the coefficient of π before finding amplitude, period, etc.
 
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  • #14
B = π not πx
 

Related to Trigonometry: Analyzing and Graphing a Cosine Function

1. What is the basic graph of cos?

The basic graph of cos is a smooth, continuous wave that starts at the point (0,1) and repeats itself every 2π units horizontally. It has a maximum value of 1 and a minimum value of -1.

2. How do you graph cos functions?

To graph a cos function, first identify the amplitude, period, and phase shift. Then, plot points on the graph by plugging in values for x and solving for y. Connect the points with a smooth curve to create the graph.

3. Can the graph of cos have a negative amplitude?

Yes, the amplitude of cos can be negative. This means that the graph will be reflected across the x-axis. The maximum value will be -1 and the minimum value will be 1.

4. What is the period of the cos graph?

The period of the cos graph is 2π. This means that the graph will repeat itself every 2π units horizontally. The formula for finding the period of a cos function is T=2π/b, where b is the coefficient of x.

5. How do you determine the range of a cos graph?

The range of a cos graph is the set of all possible y-values. Since the cosine function has a maximum value of 1 and a minimum value of -1, the range will be between -1 and 1. In interval notation, the range can be written as [-1,1].

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