SOLUTION: What is " 2y^2 - 4y -x + 12 = 0 " in the standard form of a parabola?

Question 435531: What is " 2y^2 - 4y -x + 12 = 0 " in the standard form of a parabola?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
What is " 2y^2 - 4y -x + 12 = 0 " in the standard form of a parabola?
..
2y^2 - 4y -x + 12 = 0
completing the square
2(y^2-2y+1)=x-12-2
2(y-1)^2=x-14
x=2(y-1)^2 -14
Standard form of this parabola: x=A(y-k)^2+h, with (h,k) being the (x,y) coordinates of the vertex. A is just a multiplier which affects the shape (steepness) of the curve.
Given expression is a parabola that opens rightwards with the axis of symmetry at y=1, and vertex at (-14,1). Note that this parabola is not a function since for the same x-value, you get two different values of y. See the graph of the parabola below:
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y=+-((x+14)/2)^.5+1

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