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Math Help - conics problem

  1. #1
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    conics problem

    Use 3y^2-x-12y+14=0 to do the following

    write the equation in standard form.



    B)sketch the graph(this is where i really have trouble)
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dan123 View Post
    Use 3y^2-x-12y+14=0 to do the following

    write the equation in standard form.



    B)sketch the graph(this is where i really have trouble)
    So;ve for y, start by dividin by the coefficient of the quadratic term and then completing the square.
    3y^2-x-12y+14=0\Longleftrightarrow{y^2-\frac{1}{3}x-4y+\frac{14}{3}}=0

    completing the square and doing some subratcion

    y^2-4y+4=\frac{1}{3}x-\frac{2}{3}

    You can now see that

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

    and

    y=\pm\sqrt{\frac{x-2}{3}}+2

    is that what you were looking for?
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  3. #3
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    yeah,then I just add those values for y into the graph as 2 seperate equations(+ and -) and then those are the asymptotes for the graph?
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  4. #4
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    Quote Originally Posted by dan123 View Post
    yeah,then I just add those values for y into the graph as 2 seperate equations(+ and -) and then those are the asymptotes for the graph?
    There shouldn't be any asymptotes at all...

    If you solve the equation for x instead of y, you should see that you have a quadratic in y... Quadratics don't have asymptotes...

    3y^2 - x -12y + 14 = 0

    x = 3y^2 - 12y + 14


    Notice that this is the inverse function of y = 3x^2 - 12y + 14. If you can graph this equation, apply a reflection in the line y = x.
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  5. #5
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    This is how I would do it. To me the standard form for a parabola is in the form (y - k)^2 = 4p(x - h), where (h, k) is the vertex and p is the focal length (directed distance from vertex to the focus).

    \begin{aligned}<br />
3y^2 - x - 12y + 14 &= 0 \\<br />
3y^2 - 12y &= x - 14 \\<br />
y^2 - 4y &= \frac{1}{3}x - \frac{14}{3} \\<br />
y^2 - 4y + 4 &= \frac{1}{3}x - \frac{14}{3} + 4 \\<br />
(y - 2)^2 &= \frac{1}{3}x - \frac{2}{3} \\<br />
(y - 2)^2 &= \frac{1}{3}(x - 2) \\<br />
\end{aligned}

    Because the y is squared and 4p is positive, this parabola opens to the right. The vertex is (2, 2). The axis of symmetry is y = k or y = 2. Since 4p = 1/3, p = +1/12, so the focus is at (h + p, k) or (2 + 1/12, 2) or (2.08, 2). The focal width is |4p|, or 1/3. This is the length of a chord through the parabola that crosses the axis of symmetry at the focus.

    I've attempted at attach a picture below.


    01
    Attached Thumbnails Attached Thumbnails conics problem-img_1412.jpg  
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dan123 View Post
    yeah,then I just add those values for y into the graph as 2 seperate equations(+ and -) and then those are the asymptotes for the graph?
    All you gotta do dan is realize that any time we've got a relation y=\pm{u}+c, we recognize that y=c is an axis of symetry. There fore, if you plot a few points fo the function y=+u+c and sketch a smooth curve through them. after you have a good idea of the functions behavior, you can just fold the graph paper along the line y=c, and trace. See what I mean?
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