SOLUTION: y2 - 12y + 4x + 4 =0

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Question 22081: y2 - 12y + 4x + 4 =0
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
To begin with, you can tell by looking at this that it is a parabola, since it has one variable squared but not the other. It opens either to the right or left because it has a y%5E2. You can find the vertex by completing the square.
y%5E2+-+12y+%2B+4x+%2B+4+=0
y%5E2+-+12y+=+-4x+-+4+

In order to make a perfect square trinomial on the left side of the equation, make sure you have a 1 for the coefficient of y%5E2, then take half of the -12 (which is -6) and square it (which is 36), and add + 36 to each side of the equation:
y%5E2+-+12y+______=+-+4x+-+4+______
y%5E2+-+12y+%2B36+=+-+4x+-+4+%2B+36
+%28y-6%29%5E2+=+-+4x+%2B+32

Factor out the -4 on the right side:
+%28y-6%29%5E2+=+-+4%28x+-8%29+

Because the equation is in the form y%5E2=-x, it opens to the left. To find the vertex, you must "zero out" the x and the y portions of the equation +%28y-6%29%5E2+=+-+4%28x+-8%29 which means that y = 6 and x = 8, or the vertex is at (8,6).

The graph should look like this parabola. (I don't know what is making what looks like a vertical line in the graph--that should NOT be there! Just the parabolic curve!!)

R^2 at SCC