What is the axis y=3x^2+19x-40 ?

The axis y=3x^2+19x-40 is x=-19/6

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axis y=3x^2+19x-40

axis

y = 3 x^{2} + 19 x - 40

x = - \frac{19}{6}

Solution steps

y = 3 x^{2} + 19 x - 40

Rewrite

y = 3 x^{2} + 19 x - 40

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{841}{12})) = (x - (- \frac{19}{6}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{19}{6}, - \frac{841}{12}), 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{19}{6}

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

x^{2} = - 36 y

y = + x^{2} - 12 x + 35

y = - 3 x^{2} - 6 x + 8

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