What is the axis y=12x^2+8x-15 ?

The axis y=12x^2+8x-15 is x=-1/3

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axis y=12x^2+8x-15

axis

y = 12 x^{2} + 8 x - 15

x = - \frac{1}{3}

Solution steps

y = 12 x^{2} + 8 x - 15

Rewrite

y = 12 x^{2} + 8 x - 15

in the standard form

4 \cdot \frac{1}{48} (y - (- \frac{49}{3})) = (x - (- \frac{1}{3}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{3}, - \frac{49}{3}), p = \frac{1}{48}

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}{3}

x^{2} + 12 x + y + 40 = 0

y^{2} + 4 x = 7

y = 8 x - 2 x^{2}

y = - 6 x^{2} + 600 x - 5726

y > 2 x^{2} + 5