g(x)= -x(x-1)2(x+4)2
Look at all the linear factors, They are
(-x), (x-1)2, (x+4)2
Set each one equal to 0 and solve to find the zeros:
(-x)=0, (x-1)2=0, (x+4)2=0
x=0, x-1=0, x+4=0
x=1, x=-4
x=0 is a zero of multiplicity 1 because it comes from a
(-x)=0 and (-x) is raised to the 1 power.
x=1 is a zero of multiplicity 2 because it comes from a
(x-1)2=0 and (x-1) is raised to the 2 power.
x=-4 is also a zero of multiplicity 2 because it comes from a
(x+4)2=0 and (x+4) is raised to the 2 power.
The graph will cut through the x axis at x=0 bexause it has
multiplicity 1 and since 1 is odd it will cut through.
The graph will BOUNCE OFF the x axis at x=1 bexause it has
multiplicity 2 and since 2 is even it will BOUNCE OFF the x-axis.
If you took the trouble to multiply this
g(x)= -x(x-1)2(x+4)2
all the way out, you would get
g(x) = -x5-6x4-x3+24x2-16x
and the leading term is -x5, but you can get just
the first term -x5 in your head without multiplying it
all the way out and collecting like terms like I did.
Extreme right hand and left hand behavior of graph:
Since -x5 has a negative coefficient the graph goes down on the
far right.
Since -x5 has an odd exponent the graph goes the opposite way
from the way it goes on the far right. Since it goes down on
the far right, it goes up on the far left.
So the graph looks like this:
Going from the far left going toward the right:
The graph starts from the top left and comes down from there.
That's because it goes up on the far left.
Then it comes down to -4 on the x-axis and it bounces
back up off the x-axis at x=-4. [even multiplicity]
Then it goes up and reaches a peak, then drops down to 0,
and then it cuts through the x-axis at 0 because it has odd
multiplicity.
Then it goes down and reaches a little valley.
Then it goes up to x=1 and bounces back down off the x-axis at x=1.
[even multiplicity]
Then it goes down forever on the far right.
Edwin
