What is (x+1)^2+(y+6)=19 ?

The solution to (x+1)^2+(y+6)=19 is Parabola with (h,k)=(-1,13),p=-1/4

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(x+1)^2+(y+6)=19

(x + 1)^{2} + (y + 6) = 19

(h, k) = (- 1, 13), p = - \frac{1}{4}

Solution steps

(x + 1)^{2} + (y + 6) = 19

Rewrite

(x + 1)^{2} + (y + 6) = 19

in the standard form:

4 (- \frac{1}{4}) (y - 13) = (x - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (- 1, 13), p = - \frac{1}{4}

x - y^{2} = - 5

4 x^{2} - 32 y = 0

2 x^{2} + 3 r (x) + 1 = 0

(y + 2)^{2} = 16 (x - 1)

x, (1 - t (x)) = 3 x^{2} - 4 x + 25

What is (x+1)^2+(y+6)=19 ?
The solution to (x+1)^2+(y+6)=19 is Parabola with (h,k)=(-1,13),p=-1/4