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Trigonomic Functions, Plotting from equation

  • Thread starter dylanjames
  • Start date
  • #1
24
0
Hi All,

Having a tough time with this one and i'm not sure why.
Need to state amplitude, period and phase shift of f(x)=3cos2[x-(π/4)]+1.

Amplitude being 3, period being 2π/2=π and phase shifted (π/4) to the right.
Midline would also be at y=1

Good so far?

Right, so I know that 1/4 phase would be π/4 and therefore plotting five points from the initial (π/4,4) would give me (π/2, 1), (3π/4, -2), (π, 1), (5π/4, 4)

But then plugging the initial equation into a graphing software it appears that the points do not line up.
Can anyone confirm whether this is correct?
I have been through ALL of Khan Academys videos and this lesson is driving me f'n nuts.

ANY help appreciated. Cheers
 

Answers and Replies

  • #2
12
1
Everything you posted seems to be correct.

Try typing this into the graphing software, the points should match up.

[tex] f(x) = 3cos(2x-\pi/2)+1 [/tex]
 
  • #3
33,070
4,771
Hi All,

Having a tough time with this one and i'm not sure why.
Need to state amplitude, period and phase shift of f(x)=3cos2[x-(π/4)]+1.

Amplitude being 3, period being 2π/2=π and phase shifted (π/4) to the right.
Midline would also be at y=1

Good so far?

Right, so I know that 1/4 phase would be π/4 and therefore plotting five points from the initial (π/4,4) would give me (π/2, 1), (3π/4, -2), (π, 1), (5π/4, 4)

But then plugging the initial equation into a graphing software it appears that the points do not line up.
Can anyone confirm whether this is correct?
I have been through ALL of Khan Academys videos and this lesson is driving me f'n nuts.

ANY help appreciated. Cheers
I agree with jbstemp that your work looks fine. Possibly how you typed the function into the graphing software is the problem. It probably wouldn't like 3cos2[x-(π/4)]+1, but should work with a few more parentheses, like this: 3cos(2 * (x-(π/4)))+1
 

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