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Math Help - Find parabola equation

  1. #1
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    Find parabola equation

    Tangent to the line y=12;
    x-intercepts at (-3,0) and (9,0)

    I found the vertex to be (3,12) and the parabola opens downwards, but I don't know how to go beyond this.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cityismine View Post
    Tangent to the line y=12;
    x-intercepts at (-3,0) and (9,0)

    I found the vertex to be (3,12) and the parabola opens downwards, but I don't know how to go beyond this.
    Tangent to the line y=12 where? I assume you meant that it is tangent at the vertex?

    Let f(x)=ax^2+bx+c

    Just solve the following system of equations
    f(-3)=0
    f(9)=0
    f'(3)=0
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  3. #3
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Let f(x)=ax^2+bx+c

    Just solve the following system of equations
    f(-3)=0
    f(9)=0
    f'(3)=0
    Those equations are not enough! You still need the condition that itís tangent to y=12.

    Alternative method:

    Let f(x)=k(x+3)(x-9)=k(x^2-6x-27)

    Then f'(x)=k(2x-6)=0\ \mbox{when}\ x=3

    So the curve passes through the point (3,12). Set f(3)=12 and solve for k.
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  4. #4
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    Quote Originally Posted by cityismine View Post
    Tangent to the line y=12;
    x-intercepts at (-3,0) and (9,0)

    I found the vertex to be (3,12) and the parabola opens downwards, but I don't know how to go beyond this.
    On one hand, this is supposed to be Pre-Calc. Are you allowed to use Calculus?

    On the other hand, Calculus is often used in this Pre-Calc section, so are you really alowed to use Calculus?

    If you are, then follow the replies given above, where they used f'(x) = 0 when the tangent line is horizontal.

    ----------------------
    If you are not allowed, or you are not into Calculus yet, then you can use Analytic Geometry, or the properties of a parabola, to solve the problem/question.

    So, if the horizontal line y = 12 is tangent to the parabola, and the parabola has two x-intercepts, then the y=12 line can only be tangent to the highest point, the vertex, of the parabola. The parabola being "vertical, or its axis of symmetry being vertical.

    The parabola is symmetrical about its axis of symmetry. So the x-intercepts are equidistant from the axis of symmetry.
    The absolute distance between the two x-intercepts is (3 +9) = 12 units. Half of that is 6 units. So the axis of symmetry is at x = 3.

    Hence, since the vertex is tangent to the given y = 12 line, then the vertex of the parabola is at point (3,12).

    Also, since the vertex is the highest point, then the vertical parabola opens downward.

    A standard form of a vertical parabola whose vertex is at (h,k), and that opens downward is:
    (y-k) = -a(x-h)^2 -----(i)

    Using intercept (-3,0),` and (h,k) = (3,12),
    (0 -12) = -a(-3 -3)^2
    -12 = -a(36)
    a = 12/36 = 1/3

    Therefore, the parabola is
    (y -12) = -(1/3)(x -3)^2 -----------answer.

    If you want to simplify or convert that to the form y = ax^2 +bx +c,
    y -12 = (-1/3)(x^2 -6x +9)
    y = -(1/3)(x^2) +2x -3 +12
    y = (-1/3)(x^2) +2x +9 -----------------answer also.
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