What is 4x^2+40x+106+y=0 ?

The solution to 4x^2+40x+106+y=0 is Parabola with (h,k)=(-5,-6),p=-1/16

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4x^2+40x+106+y=0

4 x^{2} + 40 x + 106 + y = 0

(h, k) = (- 5, - 6), p = - \frac{1}{16}

Solution steps

4 x^{2} + 40 x + 106 + y = 0

Rewrite

4 x^{2} + 40 x + 106 + y = 0

in the standard form

4 (- \frac{1}{16}) (y - (- 6)) = (x - (- 5))^{2}

Therefore parabola properties are:

(h, k) = (- 5, - 6), p = - \frac{1}{16}

x^{2} + y - 2 = 0

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

f (x) = x^{2} - 2 x + 5.4 f (x)

x^{2} - 10 x - 12 y - 23 = 0

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

What is 4x^2+40x+106+y=0 ?
The solution to 4x^2+40x+106+y=0 is Parabola with (h,k)=(-5,-6),p=-1/16