Results 1 to 6 of 6
Like Tree2Thanks
  • 1 Post By HallsofIvy
  • 1 Post By muddywaters

Math Help - vertex form

  1. #1
    Member
    Joined
    Feb 2013
    From
    Canada
    Posts
    131

    Post vertex form

    what is the vertex form of x2 + 5x + 6

    i dont know how to do it

    thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: vertex form

    You need to complete the square...are you familiar with this method?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2013
    From
    Canada
    Posts
    131

    Re: vertex form

    is it...

    (x^2 + 5x + 6.25)+6 - 6.25

    i do not know how to make (x^2 + 5x + 6.25) factored
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,973
    Thanks
    1638

    Re: vertex form

    Then where did you get that formula? In order to be able to do this problem, you have to know how to complete the square- you got 6.25, I presume, as (5/2)^2. Do you not know why you did that or what "completing the square" means?

    The whole point is that (x+ a)^2= x^2+ 2ax+ a^2. Comparing x^2+ 2ax with x^2+ 5x, we must have 2a= 5 or a= \frac{5}{2}= 2.5. Then a^2= (2.5)^2= 6.25. That is, (x+ 2.5)^2= x^2+ 5x+ 6.25. So adding and subtracting 6.25 from both sides, x^2+ 5x+ 6.25)+ 6- 6.25= (x+ 2.5)^2- .25 which I would have preferred to write as (x+ \frac{5}{2})^2- \frac{1}{4}.

    Now, you know (I hope!) that a square is never negative. When x= -5/2, x+ 5/2= 0 so y= (x+ 5/2)^2- 1/4 has value -1/4 while if x is any other number, (x+ 5/2)^2- 1/4 is larger than -1/4. The lowest point on the graph is (-5/2, -1/4).
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2013
    From
    Canada
    Posts
    131

    Re: vertex form

    i was actually asked to find the x and y intercepts and graph y = |x2 + 5x + 6|

    i was trying to get it into vertex form so i could get the vertex point so i know how high the curve would be i still dont know, i think x = -5/2 is right from looking at the answers but the y intercept is 6, not -1./4

    all i know is the y intercept is 6
    and x intercepts are (-3, 0) and (-2, 0)
    the vertex point is
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2012
    From
    malaysia
    Posts
    98
    Thanks
    15

    Re: vertex form

    Quote Originally Posted by Mathnood768 View Post
    i think x = -5/2 is right from looking at the answers but the y intercept is 6, not -1./4
    the mistake is that the y-intercept is not at x=-5/2, y-intercept is always at x=0
    x=-5/2 is where the vertex is, the vertex is what u get from completing the square. then the vertex is (-5/2, -1/4)

    Edit:
    but since it's a graph of y=|x^2+5x+6|, taking the absolute value, the "vertex" is (-5/2, 1/4). i think you should try to draw the graph, and you'll understand it better
    Last edited by muddywaters; July 26th 2013 at 11:58 PM.
    Thanks from Mathnood768
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: January 2nd 2013, 03:14 PM
  2. Vertex Form
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 9th 2008, 11:08 PM
  3. Changing standard form to vertex form
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: November 1st 2008, 12:25 PM
  4. Vertex form??
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 23rd 2008, 07:24 PM
  5. convert to vertex form.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 23rd 2007, 03:38 PM

Search Tags


/mathhelpforum @mathhelpforum