What is the axis x^2+8y+2x-23=0 ?

The axis x^2+8y+2x-23=0 is x=-1

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axis x^2+8y+2x-23=0

axis

x^{2} + 8 y + 2 x - 23 = 0

x = - 1

Solution steps

x^{2} + 8 y + 2 x - 23 = 0

Rewrite

x^{2} + 8 y + 2 x - 23 = 0

in the standard form

4 (- 2) (y - 3) = (x - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (- 1, 3), p = - 2

Parabola is of the form

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

and is symmetric around the

y

-axis

Axis of symmetry is a line parallel to the

y

-axis which intersects the vertex:

x = - 1

y = x^{2} - 4 x + 5

3 y^{2} = 24 x

y = x^{2} - 14 x + 45

y = - x^{2}

y = - x^{2} - 3