1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Characteristics of Polynomials

  1. Sep 12, 2016 #1
    1. The problem statement, all variables and given/known data
    I have to sketch a graph of y=x(x-3)^2

    2. Relevant equations


    3. 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. Im 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 travelling to and from?
     
  2. jcsd
  3. Sep 12, 2016 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Look at the intervals ##x < 0##, ##0< x < 3## and ##x > 3## separately. Now, think a bit: what happens to ##y## when ##x## becomes large and ##> 0##? What happens when ##x## becomes large in magnitude but ##< 0## (that is, large negative)?
     
  4. Sep 12, 2016 #3

    Mark44

    Staff: Mentor

    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.
     
  5. Sep 14, 2016 #4
    I understand it now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted