What is the vertices y^2+2y+x+5=0 ?

The vertices y^2+2y+x+5=0 is (h,k)=(-4,-1),p=-1/4

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vertices y^2+2y+x+5=0

vertices

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

(h, k) = (- 4, - 1), p = - \frac{1}{4}

Solution steps

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 4, - 1), p = - \frac{1}{4}

x^{2} + 6 x + 8

y = 2 x^{2} + 8 x - 34

f (x) = x^{2} + 6 x

3 x^{2} + 2 x + 1

3 x^{2} + 8 y = 0

What is the vertices y^2+2y+x+5=0 ?
The vertices y^2+2y+x+5=0 is (h,k)=(-4,-1),p=-1/4