Sketch the graph of y=x(x+1)(x-2)^2
First, what are the roots of x(x+1)(x-2)^2?
Roots meaning?.. Are they 0, -1, and 2? or is that something else?
@aum
Yes, -1, 0 and 2 are the roots which means the graph will cross the x-axis at the three points x = -1, x = 0 and x = 2. Next, what is the ORDER of the polynomial x(x+1)(x-2)^2 ?
Meaning i should rewrite the equation?
No. IF you were to multiply x(x+1)(x-2)^2 out what will be the highest exponent of x? (You don't have to actually multiply it out to figure out the answer).
would the highest exponent be 4?
correct. So this is a 4th degree polynomial. (or the order of the polynomial is 4).
Since 4 is even, and the coefficient of the x^4 term is POSITIVE, the END BEHAVIOR of the polynomial will be identical to that of x^2. If we were to plot x^2 what will be the end behavior?
By end behavior I mean will the left end of the graph be up or down? Will the right end of the graph be up or down?
The right side of the graph will be up and same for the left side, correct?
correct. x^2 is a parabola that opens upward. So the left end will be up and the right end will be up. Therefore, any 4th degree polynomial with positive coefficient for x^4 will have the left end up and the right end up. We just need one more piece of information. Earlier we said the roots or zeros of this polynomial are: -1, 0 and 2. What is the multiplicity of each root? In other words, how many times does each root occur in x(x+1)(x-2)^2 ?
0 only once and -1 only once, but for 2... twice? which means the line bounces? from that point?
exactly!
Since the left end is up, we start from top and draw a curve until it crosses the x-axis at x = -1, then the curve turns around and goes up through the origin (x=0). Then the curve turns down and TOUCHES the point x = 2 and without really crossing the x-axis it just turns up and keeps going up because the right end is up.
Thank you for the help and the steps through this it really helped.
You are welcome.
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