Results 1 to 9 of 9

Math Help - This parabola question baffles me!

  1. #1
    Junior Member PythagorasNeophyte's Avatar
    Joined
    Jun 2010
    Posts
    54

    This parabola question baffles me!

    A very good day to everyone here.

    I would like you to help me solve this question:
    This parabola question baffles me!-maths-parabola.jpg


    What I don't understand:
    Candidly, I don't know how to do this whole question. All I know is that the quadratic formula is y=ax^2+bx+c. But how can this formula be applied in finding p?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    433
    Thanks
    12
    You know point A is (0,8)

    Substitute x = 0, y = 8, into the equation of the graph, y = (x+p)(2-x) and solve for p. Once you have p, you can solve the rest.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member PythagorasNeophyte's Avatar
    Joined
    Jun 2010
    Posts
    54
    Hello Educated, here's my working for (a)

    <br />
y = (x+p)(2-x)

    8=(0+p)(2-0)

    8= 2p

    p= 4

    What are the coordinates of point B?
    The question doesn't state any value for x nor y. It's so tough.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    433
    Thanks
    12
    Now that you have the equation, do you know how to find the roots of an equation? To make y = 0?

    y = (x+4)(2-x)

    If x = 2, y = 0. This one is given for you, it's point c.
    If x = ..., y = 0. This one you can solve yourself.

    Solve the second root and that's the x-coordinates for b.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member PythagorasNeophyte's Avatar
    Joined
    Jun 2010
    Posts
    54
    Oh, I see, so I'll take y as 0, hence-
    <br />
0 = (x+4)(2-x)

    x+4 =0 or 2-x+0

    x=-4 or x=2

    Now, since the x value must be negative at the point B,
    I should reject x=2 and take x=-4 as the answer.

    Also, the y value at point B must be 0.

    Hence B is (-4,0).

    Now I would have to do part (c).

    The line of symmetry is the midpoint of the parabola.

    What am I suppose to do now?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    433
    Thanks
    12
    Midpoint of the parabola is halfway between the 2 roots.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member PythagorasNeophyte's Avatar
    Joined
    Jun 2010
    Posts
    54
    So is the line of symmetry equals to :

    \dfrac{-4+2}{2}

    =-1 ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    433
    Thanks
    12
    Yep. Remember that it's only the vertical line that can make it symmetrical.
    Last edited by Educated; September 22nd 2010 at 10:59 PM. Reason: Vertical
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member PythagorasNeophyte's Avatar
    Joined
    Jun 2010
    Posts
    54
    Thank you so much for your help, Educated! My problem is finally solved.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parabola Question
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 16th 2010, 05:40 AM
  2. Parabola question
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: June 6th 2010, 01:45 PM
  3. parabola question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 26th 2009, 09:44 PM
  4. Parabola Question
    Posted in the Geometry Forum
    Replies: 3
    Last Post: May 5th 2009, 10:23 AM
  5. parabola question
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 28th 2008, 08:56 AM

Search Tags


/mathhelpforum @mathhelpforum