What is the axis y^2-12x+8y+88=0 ?

The axis y^2-12x+8y+88=0 is y=-4

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axis y^2-12x+8y+88=0

axis

y^{2} - 12 x + 8 y + 88 = 0

y = - 4

Solution steps

y^{2} - 12 x + 8 y + 88 = 0

Rewrite

y^{2} - 12 x + 8 y + 88 = 0

in the standard form

4 \cdot 3 (x - 6) = (y - (- 4))^{2}

Therefore parabola properties are:

(h, k) = (6, - 4), p = 3

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 = - 4

y = + x^{2} - 7

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

2 x - y^{2} = 0

y = + 6 x^{2} + 12 x

y^{2} + 6 y + 8 x + 25 = 0