What is y^2=c^1((x+1)/(4c^1)) ?

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

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y^2=c^1((x+1)/(4c^1))

y^{2} = c^{1} (\frac{x + 1}{4 c ^{1}})

(h, k) = (- 1, 0), p = \frac{1}{16}

Solution steps

y^{2} = c^{1} \frac{x + 1}{4 c ^{1}}

Rewrite

y^{2} = c^{1} \frac{x + 1}{4 c ^{1}}

in the standard form:

4 \cdot \frac{1}{16} (x - (- 1)) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (- 1, 0), p = \frac{1}{16}

y \leq - \frac{1}{4} (x + 6)^{2} + 2

x^{2} + 8 x + 4 y + 20 = 0

y > x^{2} + 3 x

3 x^{2} - y = 5

x^{2} + 4 x - 24 y - 92 = 0

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