Discrete random variance transformation

For a discrete random variable, called X, if we know that there is another DRV which we can write as Y=aX+b - where a and b are constants:

Var(Y)=a^2\,Var(X)

We ignore the added constant and multiply the old variance by the square of a.

Why square a when finding the variance?

If we didn’t square a, we’d be finding the standard deviation (if multiplying a\times\sigma). If we want to find the variance from the standard deviation, then we’ll need to square the result, so (a\sigma)^2=a^2\,Var(X)

Why do we ignore the constant?

If we add a constant (b) to a discrete random variable, it shifts the distribution one way, but doesn’t actually change how far apart the values are. That means the variance