Results 1 to 2 of 2

Thread: turning point of a curve

  1. July 17th 2010, 05:57 AM #1
    Zharif93
    Zharif93 is offline
    Newbie Zharif93's Avatar
    Joined
    Jul 2010
    From
    KL
    Posts
    11

    turning point of a curve

    Given ( 1 , 4 ) is a turning point of the curve y = ax^3 + bx^2 - 6x - 1/2 .

    Find the values of a and b

    Follow Math Help Forum on Facebook and Google+
  2. July 17th 2010, 06:16 AM #2
    eumyang
    eumyang is offline
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    If (1, 4) is on the curve y = ax^3 + bx^2 - 6x - 1/2, plug in x and y to get
    \begin{aligned}<br /> 4 &= a(1)^3 + b(1)^2 - 6(1) - 1/2 \\<br /> 4 &= a + b - 13/2 \\<br /> a + b &= 21/2 \\<br /> \end{aligned}

    If (1, 4) is a turning point, then the derivative of y is 0 at that point:
    \begin{aligned}<br /> y' &= 3ax^2 + 2bx - 6 \\<br /> 0 &= 3a(1)^2 + 2b(1) - 6 \\<br /> 0 &= 3a + 2b - 6 \\<br /> 3a + 2b &= 6<br /> \end{aligned}

    You're left with a system of equations
    \begin{aligned}<br /> a + b &= 21/2 \\<br /> 3a + 2b &= 6<br /> \end{aligned}

    Solve for a and b.
    Follow Math Help Forum on Facebook and Google+
« Vector Form of Green's Theorem | basic definition matching. »

Similar Math Help Forum Discussions

  1. Exponential turning point q
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 9th 2010, 10:38 AM
  2. turning point problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 29th 2010, 02:09 PM
  3. Turning point
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 24th 2010, 07:24 AM
  4. Turning point of the curve
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 27th 2010, 08:55 AM
  5. Turning Point of a Parabola
    Posted in the Math Topics Forum
    Replies: 6
    Last Post: November 8th 2008, 06:33 PM

Search Tags


/mathhelpforum @mathhelpforum