What is the axis y^2+8x+2y+33=0 ?

The axis y^2+8x+2y+33=0 is y=-1

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axis y^2+8x+2y+33=0

axis

y^{2} + 8 x + 2 y + 33 = 0

y = - 1

Solution steps

y^{2} + 8 x + 2 y + 33 = 0

Rewrite

y^{2} + 8 x + 2 y + 33 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 4, - 1), p = - 2

Parabola is of the form

4 p (x - h) = (y - k)^{2}

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = - 1

y = 3 x^{2} - 6 x - 2

y = 3 x^{2} - 6 x - 1

y = (x + 4)^{2}

y = 3 x^{2} - 24 x - 15

x^{2} = 28 y