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How to find the vertex of the parabola $y^{2} + 2 y - 2 x + 5 = 0$ $y^2+2y-2x+5=0$ ?

1 Answer

Sep 11, 2015

The vertex is at $(x, y) = (2, - 1)$ $(x,y)=(2,-1)$

Explanation:

$y^{2} + 2 y - 2 x + 5 = 0$ $y^2+2y-2x+5=0$
$\to XXX x = \frac{y^{2} + 2 y + 5}{2}$ $rarrcolor(white)("XXX")x=(y^2+2y+5)/2$

Note that this is a parabola with a horizontal axis of symmetry

The vertex form for such a parabola is
$XXX x = m {(y - b)}^{2} + a$ $color(white)("XXX")x=m(y-b)^2+a$
with a vertex at $(x, y) = (a, b)$ $(x,y)=(a,b)$

Convert the equation into this form:
$x = \frac{y^{2} + 2 y + 5}{2}$ $x = (y^2+2y+5)/2$

$XXX = \frac{y^{2} + 2 y + 1}{2} + \frac{4}{2}$ $color(white)("XXX")= (y^2+2y+1)/2 + 4/2$

$XXX = \frac{1}{2} {(y + 1)}^{2} + 2$ $color(white)("XXX")=1/2(y+1)^2+2$

$XXX = \frac{1}{2} {(y - (- 1))}^{2} + 2$ $color(white)("XXX")= 1/2(y-(-1))^2+2$

which is the vertex form with $(x, y) = (2, - 1)$ $(x,y) = (2,-1)$
graph{y^2+2y-2x+5=0 [-3.57, 8.914, -4.64, 1.6]}

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