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!

Absolute parabola

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Sketch the graph of |x^2 - 2x - 3|

    2. Relevant equations

    None

    3. The attempt at a solution

    |x^2 - 2x - 3| =

    { x^2 - 2x + 3 x^2 - 2x - 3 >= 0 x >= 3 OR x >= -1
    {-x^2 + 2x + 3 x^2 - 2x - 3 < 0 x < 3 OR x < -1


    if x < -1 then y = -x^2 + 2x + 3

    if -1 <= x < 3 then y = (x^2 - 2x - 3) + (-x^2 + 2x + 3)
    = 0

    if x >= 3 then y = x^2 - 2x - 3

    I wrote this in a test and I got the part in bold wrong. I dont know what I did wrong as I have used this method of showing the positive and negative values of an absolute graph before and it has worked.
     
  2. jcsd
  3. Apr 19, 2010 #2

    Mark44

    Staff: Mentor

    The inequalities at the end of each line are wrong.
    x >= 3 OR x >= -1 is the same as saying x >= -1. By "the same as" I mean "equivalent to."
    x < 3 OR x < -1 is the same as saying x < 3.
    No, this isn't true.
    This isn't true, either.
    Look at the graph of y = x^2 - 2x - 3. This graph crosses the x-axis at (-1, 0) and (3, 0). These points determine three intervals: (-inf, -1), (-1, 3), and (3, inf). On one of these intervals the graph if y = x^2 - 2x - 3 dips below the x-axis.

    For y = |x^2 - 2x - 3|, the part that was below the x-axis is reflected across. The other two parts don't change.
     
  4. Apr 19, 2010 #3
    The funny thing is that I plotted those equations in bold (all above positive y-axis cause of absolute) and got the graph right, but I dont understand why my reasoning would be wrong as I stated the relevant intervals: x < -1, -1 <= x < 3 and x >=3.

    |x^2 - 2x - 3| =

    { x^2 - 2x + 3 x^2 - 2x - 3 >= 0 x >= 3 OR x >= -1
    {-x^2 + 2x + 3 x^2 - 2x - 3 < 0 x < 3 OR x < -1

    I got the first bold part by saying (x - 3)(x + 1) >= 0
    and the second bold part by saying (x - 3)(x + 1) < 0
    and just solving for both x values. What would be the correct inequality(s) to right?
     
  5. Apr 19, 2010 #4

    Mark44

    Staff: Mentor

    Those are the right intervals.
    I didn't notice earlier, but the first inequality in each line has extra terms in it. Probably a copy and paste error.


    The solution to this inequality is x < = -1 OR x >= 3. This represents two separate intervals.
    You have x >= 3 OR x >= -1, which as I said before is the same as saying x >= -1, which is only one interval.
    The solution to this inequality is -1 < x < 3. You had x < 3 OR x < -1, which as I said before, is the same as x < 3.

    Your problem seems to be solving quadratic inequalities. You should go back and review the section in your book that discusses this type of inequality.
     
  6. Apr 20, 2010 #5
    Oh ok I understand where I went wrong, I didnt apply the use a number line with critical values to get the interval. Thanks for the help :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Absolute parabola
  1. Distance on a Parabola (Replies: 5)

  2. Interesting Parabola (Replies: 4)

  3. Centroid of a Parabola (Replies: 3)

  4. Centroid of parabola (Replies: 13)

Loading...