Answer
The equation for the hyperbola specified in this problem is:
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
Work Step by Step
From the vertices of $(3, 0)$ and $(-3, 0)$, we already have the $a$ term for the equation. We can now plug in $3$ into the standard equation for a hyperbola to get:
$\frac{x^{2}}{3^{2}} - \frac{y^{2}}{b^{2}} = 1$
To find the value for $b$, we use the equation:
$a^{2} + b^{2} = c^{2}$
We already have the value for $a$, which is $3$, and the value for $c$, which is $5$. We plug these into the equation above to get:
$(3)^{2} + b^{2} = (5)^{2}$
We simplify this equation to get:
$9 + b^{2} = 25$
To isolate $b$, we subtract $9$ from both sides:
$b^{2} = 16$
Take the square root of both sides to get:
$b = ±4$
Now, we plug both $a$ and $b$ back into the standard equation for hyperbolas to get:
$\frac{x^{2}}{3^{2}} - \frac{y^{2}}{4^{2}} = 1$
We simplify this equation to get:
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
