graph of $y = a\cos(bx+c)+d$ 
Attached is the graph of  $y = a\cos(bx+c)+d$, where $a>0$, $b>0$, and $c>0$, and $c$ is as small as possible. Find $a + b + c + d$.


I'm having a hard time trying to solve this problem. First of all, the period seems to be $3\pi$, which means that $b=2/3$, and the amplitude, or a, is 2 and $d=1$. However, the horizontal shift of the cosine graph seems to be $3\pi/2$, which would give the value of $c=9\pi/4$. But when I add $a+b+c+d$, I get $11/3 + 9\pi/4$, which is not the correct answer. Is there a c-value smaller than the one I found? How? 

Firstly we need to find the midline of the graph, which will be the variable $d$.
Since it's maximun is $3$ and it's minimun is $-1$, the midline is:
$$y=\frac{(3)+(-1)}{2} = \fbox{d = 1}$$
We may also notice that two of the maximuns of the curve occur at: $-\frac{3\pi}{2}$ and $\frac{3\pi}{2}$, So the period of the function is $3\pi$.
Which implies:
$$\frac{2\pi}{b}=3\pi \Longrightarrow b = \frac{2}{3} $$
Now there two possible following approaches:
1) You may notice the graph to be an "upside-down" cosine graph (This will lead $c$ to be equal to $0$ which is not allowed since $c>0$).
or
2) You may notice the graph to be a shifted cosine graph
Since the nearest crest of the wave is at $-\frac{3\pi}{2}$, the original cosine function had to be moved $\frac{3\pi}{2}$ to the left (before being horizontaly scaled by $b$). So:
$$\frac{c}{b} = \frac{3\pi}{2}$$
and since $b = \frac{2}{3}$
$$c = \pi$$
Finally, since the amplitude of the graph is equal to 2
$$a=2$$
And the final equation is:
$$y=2\cos(\frac{2}{3}x+\pi)+1$$
$$a+b+c+d = 2+\frac{2}{3} + \pi+1 = \frac{11+3\pi}{3}$$
