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)
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...
Notice that this is the inverse function of . If you can graph this equation, apply a reflection in the line .
This is how I would do it. To me the standard form for a parabola is in the form , where (h, k) is the vertex and p is the focal length (directed distance from vertex to the focus).
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.