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 \cdot 2 (y - 1) = (x - 2)^{2}

Therefore parabola properties are:

(h, k) = (2, 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 = 2

y = x^{2} + 10 x

x = (- 20 y + 13)^{2}

(x - 3)^{2} = 12 (y - 1)

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

y = 10 - 2 x^{2}