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.


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

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

  • $\begingroup$ 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

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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