What is the axis y=+1/4 x^2+2/3 x-1/6 ?

The axis y=+1/4 x^2+2/3 x-1/6 is x=-4/3

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axis y=+1/4 x^2+2/3 x-1/6

axis

y = + \frac{1}{4} x^{2} + \frac{2}{3} x - \frac{1}{6}

x = - \frac{4}{3}

Solution steps

y = \frac{1}{4} x^{2} + \frac{2}{3} x - \frac{1}{6}

Rewrite

y = \frac{1}{4} x^{2} + \frac{2}{3} x - \frac{1}{6}

in the standard form

4 \cdot 1 \cdot (y - (- \frac{11}{18})) = (x - (- \frac{4}{3}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{4}{3}, - \frac{11}{18}), p = 1

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 = - \frac{4}{3}

x = + (y + 1)^{2} - 9

y = x^{2} - 3 x - 10

4 x = y^{2} - 4 y

3 x^{2} + 19 y = 0

y = 3 x^{2} + 6 x + 10