How would I graph this polynomial function...ten points for first right answer?

Question

Ok, x^3(x+2)(x-3)^2

The book tells me to
-find the zeros
-end behavior
-y-intercept
-multiplicity
-find two test points and test them

THANKS!

Answer 1

What a load of crap. You say you are going to vote 10 points for the best answer, but I look at your profile and you always let the answers go to votes. What a lying sack of ****.

Answer 2

The zeros (or x-intercepts) are at x^3 = 0, x+2 = 0 and (x -3)^2 = 0

x = 0 (multiplicity of 3)--crosses here
x = -2(multiplicity of 1)--crosses here
x = 3 (multiplicity of 2)--touch and turn

multiplyin the x^3 (x)(x^2) you get x^6, so the endbehavior is both ends go up since its a positive leading coefficient with an even degree

y-intercept : let x be 0: y = 0^3(0+2)(0-3)^2 = 0 (0,0)

Test points between the zeros: between -2 and 0 let x = -1
y = (-1)^3(-1+2)(-1-3)^2 = -16 (-1,-16)

between 0 and 3 let x = 1 and x = 2
y = (1)^3(1+2)(1-3)^2 = 12 (1,12)
y = (2)^3 (2+2)(2-3)^2 = 32 (2,32)
put zeros on the x-axis at -2, 0 and 3
Plot the points (-1, -16) and (1, 12) (2,32)
from -2 and 3 you should have the ends going upward.
then continue through -2 down to the point (-1, -16) ---curve back up through 0 and continue onto (1,12) and (2,32) ---curve back down towards the 3 on the x-axis where the graph touches and turns into the end behavior.

Answer 3

zeros 0(multiplicity 3),-2 multiplicity 1 and 3 multiplicity 2
It is of 6 degree so it ends at+infinity on both extremes
At x=0 and x=3 the graph is tangent to the x axis and at x=0
there is an inflexion point
y intercept (0,0)
two points (1,12) and (-1,-16)

Answer 4

this is the least confusing way: draw 2 columns x and y decide on user-friendly numbers for x: examples: -2, -one million, 0, one million, and 2 plug those numbers into the formulation, resolve for y, and notate the interior the y column. placed those factors on the graph and connect the dots.