What is the axis x^2-12x+16y+68=0 ?

The axis x^2-12x+16y+68=0 is x=6

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axis x^2-12x+16y+68=0

axis

x^{2} - 12 x + 16 y + 68 = 0

x = 6

Solution steps

x^{2} - 12 x + 16 y + 68 = 0

Rewrite

x^{2} - 12 x + 16 y + 68 = 0

in the standard form

4 (- 4) (y - (- 2)) = (x - 6)^{2}

Therefore parabola properties are:

(h, k) = (6, - 2), p = - 4

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 = 6

y = 2 x^{2} - 4 x

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

y = x^{2} - 10 x - 24

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

y = (x + 2)^{4}