1. ## 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)

2. Originally Posted by dan123
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.
$\displaystyle 3y^2-x-12y+14=0\Longleftrightarrow{y^2-\frac{1}{3}x-4y+\frac{14}{3}}=0$

completing the square and doing some subratcion

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

You can now see that

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

and

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

is that what you were looking for?

3. 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?

4. Originally Posted by dan123
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...

$\displaystyle 3y^2 - x -12y + 14 = 0$

$\displaystyle x = 3y^2 - 12y + 14$

Notice that this is the inverse function of $\displaystyle y = 3x^2 - 12y + 14$. If you can graph this equation, apply a reflection in the line $\displaystyle y = x$.

5. This is how I would do it. To me the standard form for a parabola is in the form $\displaystyle (y - k)^2 = 4p(x - h)$, where (h, k) is the vertex and p is the focal length (directed distance from vertex to the focus).

\displaystyle \begin{aligned} 3y^2 - x - 12y + 14 &= 0 \\ 3y^2 - 12y &= x - 14 \\ y^2 - 4y &= \frac{1}{3}x - \frac{14}{3} \\ y^2 - 4y + 4 &= \frac{1}{3}x - \frac{14}{3} + 4 \\ (y - 2)^2 &= \frac{1}{3}x - \frac{2}{3} \\ (y - 2)^2 &= \frac{1}{3}(x - 2) \\ \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

6. Originally Posted by dan123
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 $\displaystyle y=\pm{u}+c$, we recognize that $\displaystyle y=c$ is an axis of symetry. There fore, if you plot a few points fo the function $\displaystyle y=+u+c$ 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 $\displaystyle y=c$, and trace. See what I mean?