What is the vertices y^2+14y+4x+45=0 ?

The vertices y^2+14y+4x+45=0 is (h,k)=(1,-7),p=-1

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vertices y^2+14y+4x+45=0

vertices

y^{2} + 14 y + 4 x + 45 = 0

(h, k) = (1, - 7), p = - 1

Solution steps

y^{2} + 14 y + 4 x + 45 = 0

Rewrite

y^{2} + 14 y + 4 x + 45 = 0

in the standard form

4 (- 1) (x - 1) = (y - (- 7))^{2}

Therefore parabola properties are:

(h, k) = (1, - 7), p = - 1

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

x^{2} + 8 x + 12

x^{2} + 8 x + 13

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

f (x) = x^{2} - 8 x

What is the vertices y^2+14y+4x+45=0 ?
The vertices y^2+14y+4x+45=0 is (h,k)=(1,-7),p=-1