What is the axis x^2-2x-4y+13=0 ?

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axis x^2-2x-4y+13=0

axis

x^{2} - 2 x - 4 y + 13 = 0

x = 1

Solution steps

x^{2} - 2 x - 4 y + 13 = 0

Rewrite

x^{2} - 2 x - 4 y + 13 = 0

in the standard form

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

Therefore parabola properties are:

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

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

(x - 1)^{2} = - 8 (y - 4)

(2 x - 7)^{2} + 8

x = 2 y^{2} - 1 - (5 y - 6)

y = \frac{x ( 3 x - 1 )}{2}

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