How do you graph by using the zeros for #f(x)=x^2(x-4)^2#?

1 Answer
Jun 16, 2017

Answer:

#x#-intercepts: #(0, 0), (4, 0); " "y#-intercept: #(0, 0)#
touch points at #x = 0 " and " x = 4#
When #x -> oo#, #y = f(x) -> oo#.

Explanation:

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]}