Results 1 to 3 of 3

Math Help - Finding parabolic equation

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    3

    Finding parabolic equation

    Given the parabola f(x)=ax^2 we are moving it one unit on the x-axis, and two units on the y-axis. What would be the equation of the new parabola ?
    I got a hunch it would be f(x)=a(x-1)^2 + 2 but I would be grateful if someone could explain the solution and the way I approach such questions.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by leonig01 View Post
    Given the parabola f(x)=ax^2 we are moving it one unit on the x-axis, and two units on the y-axis. What would be the equation of the new parabola ?
    I got a hunch it would be f(x)=a(x-1)^2 + 2 but I would be grateful if someone could explain the solution and the way I approach such questions.
    Correct.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by leonig01 View Post
    Given the parabola f(x)=ax^2 we are moving it one unit on the x-axis, and two units on the y-axis. What would be the equation of the new parabola ?
    I got a hunch it would be f(x)=a(x-1)^2 + 2 but I would be grateful if someone could explain the solution and the way I approach such questions.
    You have got the right answer there, and for parabola graphs you follow the same methods as you would for normal:

    y=f(x-a) is a shift of plus a along the x-axis, as seen in the a(x-1)^2 part of the equation.

    y=f(x)+a is a shift of plus a along the y-axis, as seen in the + 2 part of the equation.

    The way you approached the question was fine, it is simply a question of remembering the rules

    Hope this helps explain it a bit

    Craig
    Last edited by craig; January 5th 2009 at 12:29 PM. Reason: changing spacing and layout
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 12th 2010, 09:37 PM
  2. Parabolic equation of a bridge
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 29th 2010, 05:07 AM
  3. Replies: 1
    Last Post: December 9th 2009, 08:00 PM
  4. parabolic equation graph
    Posted in the Calculus Forum
    Replies: 0
    Last Post: December 3rd 2009, 01:44 PM
  5. Finding the vertex of the parabolic graph
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 28th 2008, 03:15 PM

Search Tags


/mathhelpforum @mathhelpforum