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Math Help - Turning Points

  1. #1
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    Turning Points

    I don't know if this thread is right for the forum..
    but I have a maths assignment where it's asking to state the turning point of certain equations but i don't know how to find it.
    e.g

    y=x^2

    how to i find the turning point of that parabola?
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  2. #2
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    At the turning points of an equation the slope of y is zero. In other words the differential of the equation must be zero. So for your example:

    \frac {dy}{dx}=2x

    So we set this equal to zero to get:

    2x=0 or x=0
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  3. #3
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    Quote Originally Posted by Kiwi_Dave View Post
    At the turning points of an equation the slope of y is zero. In other words the differential of the equation must be zero. So for your example:

    \frac {dy}{dx}=2x

    So we set this equal to zero to get:

    2x=0 or x=0
    Since the question is posted in the Pre-calculus subforum I have to assume that a non-calculus approach is required.

    @OP: You're probably expected to know the turning point form of a parabola and how to get it. Certainly you should know that  y = a(x - h)^2 + k has a turning point at (h, k).
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  4. #4
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    Quote Originally Posted by amberley-jane View Post
    I don't know if this thread is right for the forum..
    but I have a maths assignment where it's asking to state the turning point of certain equations but i don't know how to find it.
    e.g

    y=x^2

    how to i find the turning point of that parabola?
    There are three approaches to finding the turning point of a parabola.

    Method 1) Due to the symmetry of the parabola, the turning point lies halfway between the x-intercepts.

    If there is only one x-intercept, then the x-intercept IS the turning point.

    So in your case, for y = x^2, the x-intercept is found by letting y = 0.

    So 0 = x^2 \implies x = 0.

    Since there is only one x-intercept, that is the x-coordinate of the turning point.

    To find the y-coordinate, substitute it into your equation.

    y = 0^2 \implies y = 0.


    So the turning point is (x, y) = (0, 0).


    For another example, look at the parabola given by the equation y = x^2 + 3x + 2.

    To find the x-intercept, let y = 0.

    So 0 = x^2 + 3x + 2

    0 = (x + 1)(x + 2)

    x + 1 = 0 or x + 2 = 0

    x = -1 or x = -2.


    To find the x-coordinate of the turning point, average the x-intercepts.

    So x_{tp} = \frac{(-1) + (-2)}{2} = -\frac{3}{2}.


    To find the y-coordinate, substitute this value into the equation.

    y_{tp} = \left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right) + 2

     = \frac{9}{4} - \frac{9}{2} + 2

     = -\frac{1}{4}.


    So the turning point is (x, y) = \left(-\frac{3}{2}, -\frac{1}{4}\right).



    Method 2)

    Method 1 is ok to use if x-intercepts exist. However, it is inappropriate to use this technique if x-intercepts do not exist.

    Therefore, most people prefer to use the "Turning Point Form" of the parabola.

    For any parabola written in the form

    y = a(x - h)^2 + k

    its turning point is at (x, y) = (h, k).

    This form of the parabola is found by completing the square.


    With your example, y = x^2, we can rewrite it as

    y = (x - 0)^2 + 0.

    So its turning point is (x, y) = (0, 0).



    With the other example, y = x^2 + 3x + 2, we complete the square to put it into turning point form.

    y = x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + 2

     = \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} + 2

     = \left[x - \left(-\frac{3}{2}\right)\right]^2 - \frac{1}{4}.


    So the turning point is (x, y) = \left( -\frac{3}{2}, \, -\frac{1}{4}\right).



    Method 3) is to use Differential Calculus, but since this is in the Pre-Calculus forum, I doubt you'd be using this method.
    Last edited by mr fantastic; August 24th 2009 at 03:38 AM. Reason: Fixed a bit of latex
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