Given: f(x) = x^2 (x-4)^2f(x)=x2(x−4)2
The zeros are the values of xx of the polynomial where f(x) = 0f(x)=0.
The xx-intercepts are the points where f(x) = 0: (x, 0)f(x)=0:(x,0)
Finding the zeros of the function:
Factor the function completely:
f(x) = x^2 (x-4)^2 " "f(x)=x2(x−4)2 Already completed
Set f(x) = 0: " "f(x) = x^2 (x-4)^2 = 0f(x)=0: f(x)=x2(x−4)2=0
x^2 = 0 " and " (x - 4)^2 = 0x2=0 and (x−4)2=0
x = 0 " and " x = 4x=0 and x=4
xx-intercepts: (0, 0), (4, 0)(0,0),(4,0)
Find the yy-intercept by setting x = 0x=0: y = 0^2(0-4)^2 = 0y=02(0−4)2=0
yy-intercept: (0, 0)(0,0)
Find multiplicity : " "2 2 at x = 0 " and " 2x=0 and 2 at x = 4x=4.
Even multiplicity causes touch points. Instead of crossing the xx-axis, the function just touches the xx-axis at the zeros. If the multiplicity is odd, the function will cross the xx-axis.
Find the limit of f(x)f(x) as x -> oox→∞ :
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]}
