What is the axis y=+(9x^2-6x+3)/3 x ?

The axis y=+(9x^2-6x+3)/3 x is x= 1/6

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axis y=+(9x^2-6x+3)/3 x

axis

y = + \frac{9 x ^{2} - 6 x + 3}{3} x

x = \frac{1}{6}

Solution steps

y = \frac{9 x ^{2} - 6 x + 3}{3} x

Rewrite

y = \frac{9 x ^{2} - 6 x + 3}{3} x

in the standard form

4 \cdot \frac{1}{12} (y - (- 3 \cdot \frac{1}{36})) = (x - \frac{1}{6})^{2}

Therefore parabola properties are:

(h, k) = (\frac{1}{6}, - 3 \cdot \frac{1}{36}), p = \frac{1}{12}

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{1}{6}

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

y = 2 x^{2} - 4 x

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

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

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