What is the vertices y^2+2y-4x+9=0 ?

The vertices y^2+2y-4x+9=0 is (h,k)=(2,-1),p=1

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vertices y^2+2y-4x+9=0

vertices

y^{2} + 2 y - 4 x + 9 = 0

(h, k) = (2, - 1), p = 1

Solution steps

y^{2} + 2 y - 4 x + 9 = 0

Rewrite

y^{2} + 2 y - 4 x + 9 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (2, - 1), p = 1

f (x) = 2 x^{2} + 5 x - 12

y = 3 x^{2} - 6 x + 3

(y + 6)^{2} = 12 (x + 1)

y = 3 x^{2} + 2 x + 6

f (x) = 4 x^{2} - 3

What is the vertices y^2+2y-4x+9=0 ?
The vertices y^2+2y-4x+9=0 is (h,k)=(2,-1),p=1