Results 1 to 5 of 5

Math Help - Writing Equations for Given Polynomial Functions and Parabolas

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    13

    Writing Equations for Given Polynomial Functions and Parabolas

    I really have no idea how to do either of those, any help would be very much appreciated. I made the graphs myself so sorry if they're a bit sloppy.

    1. ) Write one possible equation for the given polynomial functions.



    2.) Write the equation for each parabola in the form f(x)=a(x-h)^2+k

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    1) with the 3 coordinates and the local max & min, that gives you 5 conditions, so you can try a quartic equation ax^4+bx^3+cx^2+dx+e=y
    substitute the coordinates, and note that between -4 and 2, the first derivative is 0 and the second is negative. You can choose whatever point you want. Same thing for the point between 2 and 4.

    2) substitute the coordinates (-4,0) and (0,2) into a(x-h)^2+k and note that the minimum is attained at -4, so f'(-4)=0 and f''(-4)>0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Quote Originally Posted by Moo View Post
    Hello,

    1) with the 3 coordinates and the local max & min, that gives you 5 conditions, so you can try a quartic equation ax^4+bx^3+cx^2+dx+e=y
    substitute the coordinates, and note that between -4 and 2, the first derivative is 0 and the second is negative. You can choose whatever point you want. Same thing for the point between 2 and 4.
    Alternatively, could it not represent this cubic?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2008
    Posts
    13
    Quote Originally Posted by Quacky View Post
    Alternatively, could it not represent this cubic?
    The answer my teacher gave us is also on that page,

    Answers:

    1.) f(x)= a(x^3-2x^2-16x+32)

    2.) y= 1/8(x+4)^2+0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Quote Originally Posted by juvenilepunk View Post
    The answer my teacher gave us is also on that page,

    Answers:

    1.) f(x)= a(x^3-2x^2-16x+32)

    2.) y= 1/8(x+4)^2+0
    Do you understand how to get the answers? For 1), I let y=the function.
    We know that y=0 when x=2, 4 and -4
    So:
    (x-2)=0
    (x-4)=0
    (x+4)=0

    Which means that y=(x-2)(x-4)(x+4) is a possible solution, because y=0 for all of the x values provided.

    For 2), as Moo said, you've just got to understand the vertex form of a parabola.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Writing Polynomial functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 11th 2010, 07:29 PM
  2. Writing the equation for parabolas
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: January 4th 2010, 04:56 AM
  3. re-writing a quartic polynomial
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 2nd 2009, 07:40 AM
  4. Writing equations for parabolas
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 16th 2009, 11:16 AM
  5. Replies: 2
    Last Post: October 22nd 2008, 08:52 PM

Search Tags


/mathhelpforum @mathhelpforum