Question

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

Answer 1

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

Explanation:

`y^2+2y-2x+5=0`
`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
`color(white)("XXX")x=m(y-b)^2+a`
with a vertex at `(x,y)=(a,b)`

Convert the equation into this form:
`x = (y^2+2y+5)/2`

`color(white)("XXX")= (y^2+2y+1)/2 + 4/2`

`color(white)("XXX")=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)`
graph{y^2+2y-2x+5=0 [-3.57, 8.914, -4.64, 1.6]}