What is the axis x^2-2x+8y+9=0 ?

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axis x^2-2x+8y+9=0

axis

x^{2} - 2 x + 8 y + 9 = 0

x = 1

Solution steps

x^{2} - 2 x + 8 y + 9 = 0

Rewrite

x^{2} - 2 x + 8 y + 9 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (1, - 1), 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^{2} = - 8 x

y = 3 x^{2} - 12 x + 8

x^{2} = 8 y

y = x^{2} + 4 x - 45

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