What is the axis 6y=x^2-8x+14 ?

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axis 6y=x^2-8x+14

axis

6 y = x^{2} - 8 x + 14

x = 4

Solution steps

6 y = x^{2} - 8 x + 14

Rewrite

6 y = x^{2} - 8 x + 14

in the standard form

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

Therefore parabola properties are:

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

y = - x^{2} - 8

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

y = - \frac{1}{4} (x + 9)^{2} - 7

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

y^{2} = 5 x + 3