What is the vertices y^2+2x+6y+17=0 ?

The vertices y^2+2x+6y+17=0 is (h,k)=(-4,-3),p=-1/2

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vertices y^2+2x+6y+17=0

vertices

y^{2} + 2 x + 6 y + 17 = 0

(h, k) = (- 4, - 3), p = - \frac{1}{2}

Solution steps

y^{2} + 2 x + 6 y + 17 = 0

Rewrite

y^{2} + 2 x + 6 y + 17 = 0

in the standard form

4 (- \frac{1}{2}) (x - (- 4)) = (y - (- 3))^{2}

Therefore parabola properties are:

(h, k) = (- 4, - 3), p = - \frac{1}{2}

f (x) = 2 x^{2} + 3 x + 1

- (x + 2)^{2} + 6

f (x) = 2 x^{2} + 3 x - 5

y = - 4 (x - 2)^{2} + 4

4 x - y^{2} + y - \frac{31}{4} = 0

What is the vertices y^2+2x+6y+17=0 ?
The vertices y^2+2x+6y+17=0 is (h,k)=(-4,-3),p=-1/2