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

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

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(y+2)^2=16(x+2)

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

(h, k) = (- 2, - 2), p = 4

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

(h, k) = (- 2, - 2), p = 4

15 = \frac{\frac{\frac{x}{5 + R ( x )}}{R ( x )}}{15}

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

y^{2} + 16 x - 6 y - 7 = 0

\frac{f ( x )}{2 x} \leq \frac{x ^{2} + 1}{2 x}

(y + 4)^{2} = - 42 (x + 2)

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