What is the axis y=(4x^2+12x-15)/3 ?

The axis y=(4x^2+12x-15)/3 is x=-3/2

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axis y=(4x^2+12x-15)/3

axis

y = \frac{4 x ^{2} + 12 x - 15}{3}

x = - \frac{3}{2}

Solution steps

y = \frac{4 x ^{2} + 12 x - 15}{3}

Rewrite

y = \frac{4 x ^{2} + 12 x - 15}{3}

in the standard form

4 \cdot \frac{3}{16} (y - (- 8)) = (x - (- \frac{3}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{2}, - 8), p = \frac{3}{16}

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{3}{2}

(x + 2)^{2} = - 12 (y - 5)

y = x (x - 4)

y^{2} - x + 8 y + 13 = 0

y = 7 x^{2} - 9 x - 5

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