What is the axis-y^2+x+6y-14=0 ?

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axis-y^2+x+6y-14=0

axis

- y^{2} + x + 6 y - 14 = 0

y = 3

Solution steps

- y^{2} + x + 6 y - 14 = 0

Rewrite

- y^{2} + x + 6 y - 14 = 0

in the standard form

4 \cdot \frac{1}{4} (x - 5) = (y - 3)^{2}

Therefore parabola properties are:

(h, k) = (5, 3), p = \frac{1}{4}

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

y = x^{2} + x - 9

2 y^{2} - x - 12 y + 18 = 0

y = (x + 6) (x - 2)

x = (y + 2) (y - 3)

x^{2} - 6 x + 5 y = 34