Trigonometry: Analyzing and Graphing a Cosine Function

[EddieLP]

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

 

[JBA]

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?

 

[EddieLP]

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.

 

[JBA]

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

 

[Mark44]

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.

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?

 

[EddieLP]

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

 

[JBA]

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.

 

[Mark44]

Amplitude: |A| = 1/3
Period: ((2π) / |B|) = 2
Horizontal Shift: -C/B = - (3/(2π))

No, that's still wrong.

Vertical Translation: D = 0

Still wrong.

Step: ((2π)/2)/B = 1

I've made adjustments to Horizontal Shift by making it a negative number

What is B? What is C?

I put Vertical translation as 0 since there is no D in the problem.

Yes, there is.

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 $\pi$.

 

[EddieLP]

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

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?

 

[Mark44]

I reversed the order and got 2!
So Vertical Translation = 2

Yes

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))

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?

I'm trying to find out what's missing, still trying to find. any hints?

 

[JBA]

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

 

[JBA]

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.

 

[EddieLP]

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.

 

[JBA]

B = π not πx