The discussion focuses on sketching the graph of the polynomial function y=x(x-3)2. The zeros of the function are identified as 0 and 3, and the end behaviors are clarified by analyzing the polynomial's degree, which is determined to be 3 upon expansion. Participants emphasize examining the function's behavior in the intervals x < 0, 0 < x < 3, and x > 3 to understand turning points and the direction of the graph in different quadrants.
PREREQUISITESStudents studying algebra, educators teaching polynomial functions, and anyone interested in mastering graphing techniques for polynomials.
Veronica_Oles said:Homework Statement
I have to sketch a graph of y=x(x-3)^2
Homework Equations
The Attempt at a Solution
I know that the zeros are 0 and 3. The part which confuses me is that end behaviours as well as turning points. I am unsure of which way the end behaviours should be pointing. Is the highest degree 2 or 3? And how to I know which quadrants it should be traveling to and from?
I understand it now.Mark44 said:To expand on what Ray said concerning the x-intercepts, when x is "close to 0" the graph is "close to" y = x(0 - 3)2 = 9x. In other words, near x = 0, the graph of your polynomial looks a lot like the graph of the line y = 9x.
When x is "close to" 3, the graph of your polynomial resembles y = 3(x - 3)2, a parabola. I'm hopeful that you have a good idea about the shape of this parabola.
If you expand x(x - 3)2, it should be obvious what the degree of this polynomial is.