Results 1 to 3 of 3

Math Help - quadration relations

  1. #1
    Senior Member euclid2's Avatar
    Joined
    May 2008
    From
    Ottawa, Canada
    Posts
    400
    Awards
    1

    quadration relations

    given y=(x-3)(x+2)

    A) State the x-intercepts
    B) State the x-coordinate of the vertex
    C) Calculate the y-coordinate of the vertex
    D) Sketch the relation
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    A) The x-intercepts are the points the graph cross when y = 0. So if
    0 = (x-3)(x+2)

    What are the x-intercepts now?

    B and C) Okay, here you're gonna have to multiply both expressions, then complete the square. Do you know how to complete the square? In the end, you'll get an expression (after completing the square) like this:
    y = a(x-h)^2 + k

    The vertex is (x,y) = (h,k)

    Be careful! If what you get in the end is a(x+h)^2 + k remember that it's actually  a(x-(-h))^2+ k, so the vertex (x,y) = (-h, k)

    D) The relation is a parabola. You got the x-intercepts and the vertex, now plot them and sketch a parabola that contains these points.

    If there is something you don't understand, don't hesitate to say so.
    Last edited by Chop Suey; August 8th 2008 at 12:01 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2007
    From
    Melbourne
    Posts
    428
    You can also find the x-coordinate of the vertex by using the fact that it is half-way between the x-intercepts. The y-coordinate can then be found by substituting in the x-coordinate.

    Edit:typo in my typo correction :S
    By the way, there is a typo in Chop-Suey's post; after completing the square the expression should be in the form y = a(x-h)^2+k
    Last edited by badgerigar; August 7th 2008 at 11:52 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relations and Functions - Inverse Relations Question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 13th 2011, 12:20 PM
  2. Replies: 1
    Last Post: September 19th 2011, 01:09 PM
  3. Relations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 27th 2010, 06:47 AM
  4. Relations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 26th 2010, 09:52 PM
  5. help with relations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: December 6th 2008, 03:56 PM

Search Tags


/mathhelpforum @mathhelpforum