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cos graph

This graph is supposed to be of form $a\cos(bx+c)+d$. I'm pretty sure that $a$, the amplitude, is $|2|$ and $b$, the period ($\frac{2\pi}{b}$), is $\frac{2}{3}$ (though some confirmation would be nice).

However, I am not sure about the phase shift $(-\frac{c}{b})$ and also the vertical shift $(d)$. It seems odd to me how the lowest point pictured on the graph is $-1$ and the highest point is $4$. Would I just shift the whole graph down by $1$ to even things out?

I looked at the graph $2\cos(\frac{2}{3}x)$ on WolframAlpha:http://www.wolframalpha.com/input/?i=graph+2cos%28%5Cfrac%7B2%7D%7B3%7Dx, and the resulting graph looks like the graph pictured above reflected over the $x$ axis, so I am not sure what to do. Help would be appreciated.

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An easy way to find the vertical shift is to find the average of the maximum and the minimum. For cosine that is zero, but for your graph it is $\frac{-1+3}2=1$. Therefore the vertical shift, $d$, is $1$.

Notice that the amplitude is the maximum minus the average (or the average minus the minimum: the same thing). In your graph it is $3-1=2$ (or $1--1=2$), as you already knew. This gives us a check on both the vertical shift and the amplitude. By the way, $a$ could be the negative of the amplitude, though it is usually taken to be the amplitude.

The period is $\frac p{|b|}$, where $p$ is the period of the "base" function. The period of the graph is seen to be $3\pi$ and cosine's period is $2\pi$, so a positive value for $b$ is $\frac{2\pi}{3\pi}=\frac 23$. Note the period is not $b$ as you wrote. Again, $b$ could be negative but it is usually taken to be positive.

An easy way to find the phase shift for a cosine curve is to look at the $x$ value of the maximum point. For cosine it is zero, but for your graph it is $3\pi/2$. That is your phase shift (though you could also use $-3\pi/2$). By the way, the formula for phase shift is not $c$, but $-\frac cb$ to the right. This is easier to see if you rewrite the formula as

$$f(x)-d=a\cos\left[b\left(x--\frac cb \right)\right]$$

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  • $\begingroup$ Thank you for the advice on the phase shift of cos...it was very helpful. I ended up getting $y = 2cos(\frac{2}{3}x+\pi)+1$. Did i do that correctly? a, b, and c need to be positive and c needs to be the smallest positive number, which I believe is $\pi$. $\endgroup$ – Mathy Person Jan 6 '15 at 18:58
  • $\begingroup$ @MathyPerson: Yes, that is correct. Note that in this formula the phase shift is $-3\pi/2$: negative, which is necessary if both $b$ and $c$ are positive. $\endgroup$ – Rory Daulton Jan 6 '15 at 19:01
  • $\begingroup$ Oh, I see. Originally I had $y=2cos(\frac{2}{3}x-\pi)+1$, but then $c = -\pi$, which is negative, so I just used $\pi$ instead. $\endgroup$ – Mathy Person Jan 6 '15 at 19:03
  • $\begingroup$ You made a typographical error at the end. You meant to write $-\frac{c}{b}$ rather than $-\frac{b}{c}$ in the formula on the last line. $\endgroup$ – N. F. Taussig Apr 14 '17 at 18:50
  • $\begingroup$ @N.F.Taussig: Yes, you are right. I have corrected it now. Thanks for the heads-up. $\endgroup$ – Rory Daulton Apr 14 '17 at 19:29
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To figure out the vertical shift, what would you do to a function centered on the x axis to achieve the given graph. In this case you are adding 1. (going from a function with a range of $[-2,2]$ to $[-1,3]$)

Then a simple way to find a phase shift is to look at the part of the graph that would normally correspond to $\cos(0)$. In this case, it is the maximum at $x=\frac{3\pi}{2}$. So, what can we do to $\cos{(b\frac{3\pi}{2})}$ to make it transform to $\cos(0)$.

So in this case, the way you have the function written, $c$ should be $-b\frac{3\pi}{2}$ or $-\pi$. Note in this case $\pi$ also works since it is symmetric on the y axis.

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