What is (x+3)^2+(y-0)=16 ?

The solution to (x+3)^2+(y-0)=16 is Parabola with (h,k)=(-3,16),p=-1/4

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(x+3)^2+(y-0)=16

(x + 3)^{2} + (y - 0) = 16

(h, k) = (- 3, 16), p = - \frac{1}{4}

Solution steps

(x + 3)^{2} + (y - 0) = 16

Rewrite

(x + 3)^{2} + (y - 0) = 16

in the standard form:

4 (- \frac{1}{4}) (y - 16) = (x - (- 3))^{2}

Therefore parabola properties are:

(h, k) = (- 3, 16), p = - \frac{1}{4}

y + 16 = x^{2} + 8

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

3 y^{2} - 2 (4 + 5 x) = 2 y^{2} + x

(x - 1)^{2} = - 8 (y + 2)

x = (y + 6)^{2}

What is (x+3)^2+(y-0)=16 ?
The solution to (x+3)^2+(y-0)=16 is Parabola with (h,k)=(-3,16),p=-1/4