What is the axis y=3x^2+5x-4 ?

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axis y=3x^2+5x-4

axis

y = 3 x^{2} + 5 x - 4

x = - \frac{5}{6}

Solution steps

y = 3 x^{2} + 5 x - 4

Rewrite

y = 3 x^{2} + 5 x - 4

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{73}{12})) = (x - (- \frac{5}{6}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{5}{6}, - \frac{73}{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{5}{6}

y = 9 x^{2} - 5 x

y = (x - 7)^{2} + 10

6 x^{2} - 8

y = x^{2} - 6 x - 12

(x + 9)^{2} + 2 (y - 4) = 0