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

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

axis

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

x = - 1

Solution steps

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 1, - 4), 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 = - 1

y = 7 x^{2} + 5

y = - x^{2} + 8 x

y = - 16 x^{2} + 40 x + 2

x = - 2 y + y (y - 2) - 2 y (y - 1) - (y - 3)

y = - 0.5 x^{2} + 5.5