You can factor:
#f(x) = 1/(x(x-4))#
You know that #1/0# is undefined. Therefore, you know that there are asymptotes at #x = 0# and #x = 4#. However, there are values defined at #0 < x < 4#. You can tell that if you plug in values from #0# to #4# non-inclusive, #f(x)# will be negative.
You can deduce from plugging in a few important points (e.g. #x = 1, 2, 3#) that it's basically somewhat like a parabola that embraces the two asymptotes.
The form of #1/x^2# as-written is like flipping Quadrant III of #1/x# over the x-axis, and making the curve more "squished" inwards towards the origin.
I would expect then for the graph to be a combination of an upside-down "parabola"-like shape, plus the two halves of #1/x^2# on either side of it, with vertical asymptotes at #x = 0, 4#.
graph{1/(x^2 - 4x) [-10, 10, -5, 5]}
