Results 1 to 4 of 4

Math Help - Find a

  1. #1
    Banned
    Joined
    Jun 2007
    From
    Jackson, New Jersey
    Posts
    37

    Find a

    If f(x) = 3x^2 - 7 and g(x) = 2x + a, find a so that the graph of f(g(x)) crosses the y-axis at 68.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by blueridge View Post
    If f(x) = 3x^2 - 7 and g(x) = 2x + a, find a so that the graph of f(g(x)) crosses the y-axis at 68.
    f(g(x)) = 3(2x+a)^2 - 7

    The y-intercept is when x=0.
    So when x=0 we have f(g(0)) = 68
    That means,
    3(2\cdot 0 + a)^2 - 7 = 68
    Thus,
    3(2a)^2 - 7 = 68

    3(2a)^2 = 75 \Rightarrow (2a)^2 = 25
    Thus,
    2a = \pm 5
    Thus,
    a= \pm 2.5
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Jun 2007
    From
    Jackson, New Jersey
    Posts
    37

    curious...

    I'm curious, but what would be the answer if g(f(x)) crosses the x-axis at 68?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    Quote Originally Posted by blueridge View Post
    I'm curious, but what would be the answer if g(f(x)) crosses the x-axis at 68?
    So you're trying to find: g(f(68))
    Since x = 68

    First find g(f(x)):.. 2(3x^2 - 7) + a.. \Rightarrow.. 6x^2 - 14 + a

    Since it crosses the x - axis at 68, the y coordinate is zero.

    g(f(x)) is the same as the y coordinate:.. 0 = 6(68)^2 - 14 + a.. \Rightarrow.. a = -27730
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: March 22nd 2011, 05:57 PM
  2. Replies: 2
    Last Post: July 5th 2010, 09:48 PM
  3. Replies: 1
    Last Post: February 17th 2010, 04:58 PM
  4. Replies: 0
    Last Post: June 16th 2009, 01:43 PM
  5. Replies: 2
    Last Post: April 6th 2009, 09:57 PM

/mathhelpforum @mathhelpforum