Explain how to get the vertical values when $n = 10$, $p = .5$, $\mu = 5$ and $\sigma^2 = 2.5$ 

There's a bit of confusion in your question so here's some quick explanations. You can "predict the shape of a binomial probability distribution just knowing $n$ and $p$" becauase $n$ and $p$ are the only parameters in a binomial distribution. Specifically, the binomial distribution is given a total number of trials $n$ and a probability of success $p$, the probability of $x$ successes is given by:

$$p(X = x) = {n\choose x}p^x (1 - p)^{n-x} $$

This is what gives you the "values of the vertical axis", for a given $n$ and $p$, you pick a point on the horizontal axis, $x$, and plug it into the above formula to get the probability of it occurring and that will be the vertical axis, representing the probaiblity.

The point you make about $\mu$ and $\sigma^2$ coming into play is a different topic, that it when you are trying to approximate the binomial distribution using a normal distribution. A normal distribution is the standard bell curve and a commonly used distribution in probability and statistics. It so happens that when the binomial distribution has a large number of trials, that is, $n$ is high, it can be very well approximated by the normal distribution using the substitutions that you mention.