Answer
$horizontal$, $(-a,0)$, $(a,0)$. $c=\sqrt {a^a+b^2}$, $(-4,0)$, $(4,0)$, $(-5,0)$, $(5,0)$
Work Step by Step
1. For a hyperbola with equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, a\gt0, b\gt0$, the transverse axis will be $horizontal$ (in contrast to equation $-\frac{x^2}{b^2}+\frac{y^2}{a^2}=1, a\gt0, b\gt0$).
2. In this case, the vertices will be at $(-a,0)$ and $(a,0)$.
3. For the foci $(\pm c, 0)$, we can used a formula $c=\sqrt {a^a+b^2}$ to calculate its value.
4. With the example of $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$, its vertices will be at $(-4,0)$ and $(4,0)$, as $c=\sqrt {4^2+3^2}=5$, the foci will be at $(-5,0)$ and $(5,0)$
