I don't understand why this graph is a sine function. My book said that a cosine curve has a y-intercept at one of the high or low points while a sine curve has one that's in the middle curve. The y-intercept on this curve is at the high point isn't it? So, why isn't it a cosine curve?

 A: "My book said that a cosine curve has a y-intercept at one of the high or low points while a sine curve has one that's in the middle curve."
Your book is wrong.  sine curves and cosine curves are the same thing as $\sin (kx) = \cos (kx - \frac {\pi}2) $.
But a function of the form $y = k\sin(jx) + c$ will have the $y$ intercept in the middle while a function of the form $y = k\cos(jx) +c$ will have the $y$ intercept in the extreme.
Likewise a function of the form $y = k\sin(jx - d) + c$ will have a value at $x = d$ of the middle while $y = k\cos(jx - d) + c$ will have a value at $x = d$ and the extreme.  (basically the entire graph will have been shifted to the right by $d$ units).
  " The y-intercept on this curve is at the high point isn't it?"
Yes, you are correct.  
"So, why isn't it a cosine curve?"
Since your book is wrong this isn't the right question. The right question is
"So, why isn't this of the form $y=k\cos(jx)+ c$?"
It is.  Your book is still wrong.
You have a truly crappy book. 
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the graph of $y = k \text{function}(jx + d) + c = k\text{function}(j(x + \frac dj))$ will be the same as the graph $y = \text{function}(x)$ with:
i) the graph will be shifted over to the left $d$ units.
ii) centering at $x = -d$ the graph will be made $j$ times thinner.
iii) centering at $y = 0$ the graph will be stretch $k$ times larger.
iv) the graph will be shifted up $c$ units.
(Alternatively i) and ii) can be altered that is is made thinner centered at $x = 0$ and shifted to the left by $\frac dj$ units$.)
Using this, we know that because $\cos (x) = \sin (x + 90^{\circ})$, the have the same graph but shifted to the right by $90^{\circ}$.  (Which is just far enough to shift the $y$-intecept from being in the middle $(\sin (0) = 0)$ to the highest value ($\sin (0 + 90^{\circ}) = 1$).)
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A correct description of that graph could be any one of the following:
a) $y = 2\cos (3x) - 1$
b) $y = 2\sin (3x + 90) -1$
c) $y = 2\sin (3(x+30)) - 1$
d) $y = 2\sin (3(x - 90)) - 1$.
A: None of the given options match the graph shown, as $\sin(0)=0$ implies that $g(0)$ is either $-1$, $-2$, or $-3$ depending on which answer choice you investigate. So either the question is in error, or there is information relevant to the question that doesn't appear in the picture.
However, it is possible to use sine to define that curve, but it would have to be shifted horizontally. For instance, $\cos(x) = \sin(x+90^\circ)$. It is likely that you will see these horizontal translations in the future.
