Q:

Find equation for hyperbola

Just taking (failing) a simple algebra class, can't figure this one out and no one can explain it to me and the book just tells me to do it.

Find an equation for the hyperbola described:
foci at $(-4,0)$ and $(4,0)$; asymptote the line $y=-x$.

So I know that since the numbers on the $x$ axis are changing it will be a horizontal hyperbola. That means $0$ is the center and $c$ is $4$.
I know the slope is $b/a$ for horizontal equations so I know that $b/a = -1$
From that I can get $b = -a$.
This is as far as I can get, my book basically does these steps in the solution manual except they get $-b/a = -1$,   $b=a$   I don't even know why. I can't work past this point without graphing and I know there is suppose to be a way just by working out the algebra but I don't see a solution. 
I might not be prepared for this test but I am prepared to fail the test. 

A:

So, you know that the equation is going to look like
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$$
You already figured out that $a=-b$. So now you need to figure out what values of $a$ and $b$ will make the foci be at the given point $(4,0)$ and $(-4,0)$. 
Since you know that $a=-b$, then you know that $a^2=b^2$, so you can rewrite the equation as
$$\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$$
or, by clearing denominators (multiplying through by $a^2$)
$$x^2 - y^2 = a^2.$$
Now: where does the hyperbola intersect the $x$ axis? If you have a point $(X,0)$ in the hyperbola, then plugging it in you will have
$$X^2 = a^2.$$
So if you can determine where the hyperbola intersects the $x$-axis, you can use that to figure out the value of $a$, and so the value of $b$, and so the equation for the hyperbola. 
Note: There are many, many, many ways to finish the problem, depending on what you know about hyperbolas in general (which is precisely why I asked you in comments what you knew). 
Added. One possible way to finish the problem is to use the  eccentricity of the hyperbola. The distance from the center to the foci is $a\varepsilon$, where $\varepsilon$ is the eccentricity of the hyperbola. The eccentricity is always equal to
$$\varepsilon = \sqrt{ 1 + \frac{b^2}{a^2}}$$
so here, since $a=b$, that means that the eccentricity is $\sqrt{2}$. Therefore, since the distance from the center to the foci is $4$, this tells you that $4 = a\varepsilon = a\sqrt{2}$. You now know what $a$ is, and thus, you know what $a^2=b^2$ is, and so you know the equation of the hyperbola. 
Added 2. Another possibility you discuss in the comments. You have $a$ is the distance from the center to the vertices; $b$ is the distance from the center to the conjugate axis. These distances satisfy $c^2 = a^2+b^2$, where $c$ is the value $4$ that you have. Since $a^2=b^2$, then you have $c^2=2a^2$, or $c = \sqrt{2a^2} = \sqrt{2}|a| = a\sqrt{2}$. You already know what $c$ is, so now you can solve for $a$. Once you know $a$, you also know $b$ (since $a=-b$) which gives you the information you need.

A:

As the foci are on the $x$-axis and symmetric with respect to the $y$-axis,
the equation of the hyperbola is
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1.$$
Such an hyperbola has two asymptotes: 
$$y=\frac{b}{a}x\qquad\text{and}\qquad y=-\frac{b}{a}x,$$ where both $a$ and $b$ are positive. The given asymptote is $y=-x$ and the
other one is $y=x$. The equation $y=-\frac{b}{a}x$ should be equivalent to $
y=-x$, which implies that $-\frac{b}{a}=-1$. Hence $b=a$. Now you use the
information on the foci to find $a$. The distance from each focus to the
origin is $c=4$. These numbers are related by the equation $$a^{2}+b^{2}=c^{2}
.$$ For $b=a$ and $c=4$, we have $a^{2}+a^{2}=16$, thus $a=2\sqrt{2}$ (the
other solution is negative), and the equation of the hyperbola is
$$\frac{x^{2}}{8}-\frac{y^{2}}{8}=1.$$

