How do you find the vertex and axis of symmetry, and then graph the parabola given by: #f(x)=(x-5)^2 - 9#?

There are 2 methods:
Method 1
Multiply out and write in standard quadratic form #y=ax^2+bx+c#
Then vertex (max/min) value occurs at point when derivative is zero, ie. when #2ax+b=0=>x=(-b)/(2a)#
So in this case, we get :
#y=f(x)=x^2-10x+16#
#therefore dy/dx=0iff2x-10=0iffx=10/2=5#
#therefore f(5)=-9=># vertex occurs at #(5;-9)#

Method 2
The completed square form of the quadratic function is #y=a(x-p)^2+q#
In this case, #(p;q)# represents the vertex, ie p is the x value at the axis of symmetry and q is the corresponding y value.
So in this example, p = 5 and q = - 9 and so we immediately get the vertex or turning point at #(5;-9)#

Graphically :

graph{((x-5)^2)-9 [-32.47, 32.47, -16.24, 16.25]}

That equation is written in the vertex form #f(x)=a(x-h)^2+k#.

Vertex
The vertex is the point #(h,k)#. Take a look at the equation. You'll see that your #h# is #5#, and your #k# is #-9#. Therefore, your vertex is #(5,-9)#. Plot that on your graph.

Axis of Symmetry
In a quadratic function, the axis of symmetry is a vertical line that passes through the vertex. The axis of symmetry is written as #x=h#. Looking at your equation, #h# is equal to #5#. Therefore, the axis of symmetry is #x=5#. You don't actually have to draw this on your graph though.

Finally, when drawing the graph, create a table of #x# and #y# values. To fill it up, substitute any value to #x# and write down the corresponding value of #y#. Try to substitute easy numbers like 0, 1, and 2. After that, plot the points and draw the parabola. It should look like this:
graph{(x-5)^2-9 [-20, 20, -10, 10]}