I am trying to undersant my professors notes. What I don't understand is how we can see that this is a finite impulse response filter by looking at the Z-transform or the graphical representation:
I know that a FIR filter has an impulse response that goes towards zero when discrete time goes towards infinity. I don't know what is meant by "does not have any denominator" comment.
An FIR or Finite Impulse Response filter has a finite number of coefficients. The coefficients don't go to zero in the limit. The coefficients of filter $h[k]$ are actually zero for $k \geq K$, assuming $h$ is an FIR filter of length $k$.
Now consider taking z-transform of $h$. By definition:
$H(z) = \sum_{k=-\infty}^{\infty} h[k]z^{-k}$ = $\sum_{k=0}^{K-1} h[k]z^{-k}$.
The second equality is critical and it follows because only $K$ of the filter coefficients are non-zero (FIR filter).
Now you can see that $H(z)$ is of the form $h[0] + h[1]z^{-1} + h[2]z^{-2} + ... + h[K-1]z^{-K+1}$, i.e. it is just a polynomial in $z^{-1}$ without a denominator.