What is (2y+3)^2=-16(2x+1) ?

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

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(2y+3)^2=-16(2x+1)

(2 y + 3)^{2} = - 16 (2 x + 1)

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

Solution steps

(2 y + 3)^{2} = - 16 (2 x + 1)

Rewrite

(2 y + 3)^{2} = - 16 (2 x + 1)

in the standard form:

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

Therefore parabola properties are:

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

y^{2} - 4 y + x = 0

x^{2} - 7 x = y - 8

x = \frac{y ^{2}}{2} - \frac{1}{2}

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

x, Y (x)^{2} = - 32 x

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