SOLUTION: Given the equation 3x^2-42x+6y+147=0 find vertex, if parabola opens (up,down,left, or right), and focus

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Question 435039: Given the equation 3x^2-42x+6y+147=0
find vertex, if parabola opens (up,down,left, or right), and focus
Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
Transform the equation 3x%5E2-42x%2B6y%2B147=0 in the standard form,
%28X-Xo%29%5E2=4c%28Y-Yo%29,
3%28x%5E2-14x%2B49%29%2B6y%2B147-3%2A49=0 => 3%28x-7%29%5E2=-6y => %28x-7%29%5E2=-2%28y-0%29
From the last expression of the parabola,s equation we have: the vertex is the
point (7, 0) and the focus will be the point: [4c=-2 => c=-1/2],(7, -1/2).
Done.