What is -4z^2+2y-7=0 ?

The solution to -4z^2+2y-7=0 is Parabola with (h,k)=(0, 7/2),p= 1/8

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-4z^2+2y-7=0

- 4 z^{2} + 2 y - 7 = 0

(h, k) = (0, \frac{7}{2}), p = \frac{1}{8}

Solution steps

- 4 z^{2} + 2 y - 7 = 0

Rewrite

- 4 z^{2} + 2 y - 7 = 0

in the standard form:

4 \cdot \frac{1}{8} (y - \frac{7}{2}) = (z - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, \frac{7}{2}), p = \frac{1}{8}

(y - 5)^{2} = - 12 (x - 3)

f, f (x)^{- 1} (x) = (\frac{1}{x - 2})

y^{2} - 16 x - 2 y + 49 = 0

4 f (x) = 7 x^{2} - 12 x + 15

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

What is -4z^2+2y-7=0 ?
The solution to -4z^2+2y-7=0 is Parabola with (h,k)=(0, 7/2),p= 1/8