What is the axis x^2-6x+8y-31=0 ?

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axis x^2-6x+8y-31=0

axis

x^{2} - 6 x + 8 y - 31 = 0

x = 3

Solution steps

x^{2} - 6 x + 8 y - 31 = 0

Rewrite

x^{2} - 6 x + 8 y - 31 = 0

in the standard form

4 (- 2) (y - 5) = (x - 3)^{2}

Therefore parabola properties are:

(h, k) = (3, 5), 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 = 3

y = - 4 x^{2} - 8 x - 6

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

x^{2} = - 12 y

3 x^{2} + 8 y = 0

y = 3 x^{2} + 9 x + 9