What is the axis y=-1/6 x^2+4x+10 ?

The axis y=-1/6 x^2+4x+10 is x=12

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axis y=-1/6 x^2+4x+10

axis

y = - \frac{1}{6} x^{2} + 4 x + 10

x = 12

Solution steps

y = - \frac{1}{6} x^{2} + 4 x + 10

Rewrite

y = - \frac{1}{6} x^{2} + 4 x + 10

in the standard form

4 (- \frac{3}{2}) (y - 34) = (x - 12)^{2}

Therefore parabola properties are:

(h, k) = (12, 34), 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 = 12

f (x) = - (x - 9)^{2} - 30

x^{2} - 4 x - 5

y = 5 \cdot (x + 7 + 149) (x + 7 - 149)

30 x^{2} + 90 x + y + 64 = 0

y = (x - 7) (x - 9)