Given: #f(x) = x^2 (x-4)^2#
The zeros are the values of #x# of the polynomial where #f(x) = 0#.
The #x#-intercepts are the points where #f(x) = 0: (x, 0)#
Finding the zeros of the function:
Factor the function completely:
#f(x) = x^2 (x-4)^2 " "# Already completed
Set #f(x) = 0: " "f(x) = x^2 (x-4)^2 = 0#
#x^2 = 0 " and " (x - 4)^2 = 0#
#x = 0 " and " x = 4#
#x#-intercepts: #(0, 0), (4, 0)#
Find the #y#-intercept by setting #x = 0#: #y = 0^2(0-4)^2 = 0#
#y#-intercept: #(0, 0)#
Find multiplicity : #" "2# at #x = 0 " and " 2# at #x = 4#.
Even multiplicity causes touch points. Instead of crossing the #x#-axis, the function just touches the #x#-axis at the zeros. If the multiplicity is odd, the function will cross the #x#-axis.
Find the limit of #f(x)# as #x -> oo# :
If you distribute #f(x) = x^4 + ...#. When #x -> oo#, #y = f(x) -> oo^4 = oo#.
This means the graph starts high and to the right, touches the #x#-axis at #x = 4#, increases in the positive #y#-direction for a short distance, then eventually decreases and touches the #x#-axis at #x = 0# and then increases without bound.
The graph looks like a W:
graph{x^2(x-4)^2 [-43.17, 36.83, -5.76, 34.24]}
