SOLUTION: find the vertex, focus, and directrix. Then draw the graph 1), {{{x^2+6x=8y-1}}} 2), {{{y=(1/16)*(x+1)^2 -2}}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the vertex, focus, and directrix. Then draw the graph 1), {{{x^2+6x=8y-1}}} 2), {{{y=(1/16)*(x+1)^2 -2}}}      Log On


   

Question 732375: find the vertex, focus, and directrix. Then draw the graph
1), x%5E2%2B6x=8y-1
2), y=%281%2F16%29%2A%28x%2B1%29%5E2+-2
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Here is a bit of help on your question number 2.

When you derive the equation for a general parabola through distance formula and directrix and focus, you get a result of y=4p%2Ax%5E2, and p is the distance from the vertex to the focus and it is also the distance from the vertex to the directrix.

The shape of your parabola in #2 is the same shape as y=%281%2F16%29x%5E2, only the position has changed. Compare this with the general equation for the untranslated equation for a parabola. You can get the value of p through equating 4p with (1/16). 4p=%281%2F16%29.

As for the vertex, look for the information from the given equation (which is already given in standard form) to find the "(h, k)" point.
You have y=%281%2F16%29%28x-%28-1%29%29%5E2-2, so your vertex is (-1, -2).
If you did not yet find p, do it NOW. You are ready to find the focus [(-1, -2+p)] and the directrix.