What is the vertex of $y^{2} - 2 y - 2 x + 5 = 0$ $y^2-2y-2x+5=0$ ?

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Answer 1 · 2018-07-04T21:52:00

$(2, 1)$ $(2, 1)$

Given equation:

$y^{2} - 2 y - 2 x + 5 = 0$ $y^2-2y-2x+5=0$

$y^{2} - 2 y + 1 - 1 - 2 x + 5 = 0$ $y^2-2y+1-1-2x+5=0$

${(y - 1)}^{2} - 2 x + 4 = 0$ $(y-1)^2-2x+4=0$

${(y - 1)}^{2} = 2 x - 4$ $(y-1)^2=2x-4$

${(y - 1)}^{2} = 2 (x - 2)$ $(y-1)^2=2(x-2)$

Above is the equation of horizontal parabola: $Y^{2} = 4 a X$ $Y^2=4aX$ which has

Vertex: $(X = 0, Y = 0) \equiv (x - 2 = 0, y - 1 = 0) \equiv (2, 1)$ $(X=0, Y=0)\equiv(x-2=0, y-1=0)\equiv(2, 1)$

What is the vertex of y^2-2y-2x+5=0y2−2y−2x+5=0 y^2-2y-2x+5=0?