What is the axis 3x^2+19y=0 ?

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axis 3x^2+19y=0

axis

3 x^{2} + 19 y = 0

x = 0

Solution steps

3 x^{2} + 19 y = 0

Rewrite

3 x^{2} + 19 y = 0

in the standard form:

4 (- \frac{19}{12}) (y - 0) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = - \frac{19}{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 = 0

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

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

y = 10 (x - 6)^{2} + 5

y = 11 x - x^{2}

y = - \frac{1}{2} (x + 6)^{2} + 8