What is the axis x^2-4x+8y-12=0 ?

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

axis

x^{2} - 4 x + 8 y - 12 = 0

x = 2

Solution steps

x^{2} - 4 x + 8 y - 12 = 0

Rewrite

x^{2} - 4 x + 8 y - 12 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (2, 2), 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 = 2

8 (x - 2) = (y + 3)^{2}

6 y = x^{2} - 8 x + 14

y = - x^{2} - 8

y = (2 x + 3)^{2}

y = - \frac{1}{4} (x + 9)^{2} - 7