What is the axis y^2-4x+6y+17=0 ?

The axis y^2-4x+6y+17=0 is y=-3

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axis y^2-4x+6y+17=0

axis

y^{2} - 4 x + 6 y + 17 = 0

y = - 3

Solution steps

y^{2} - 4 x + 6 y + 17 = 0

Rewrite

y^{2} - 4 x + 6 y + 17 = 0

in the standard form

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

Therefore parabola properties are:

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

Parabola is of the form

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

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = - 3

x = (y - 2) (y + 1)

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

y^{2} - 12 x + 20 = 0

y = - 3 x^{2} - 2 x

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