What is the axis-4y^2-12y+1=3x-5 ?

The axis-4y^2-12y+1=3x-5 is y=-3/2

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axis-4y^2-12y+1=3x-5

axis

- 4 y^{2} - 12 y + 1 = 3 x - 5

y = - \frac{3}{2}

Solution steps

- 4 y^{2} - 12 y + 1 = 3 x - 5

Rewrite

- 4 y^{2} - 12 y + 1 = 3 x - 5

in the standard form

4 (- \frac{3}{16}) (x - 5) = (y - (- \frac{3}{2}))^{2}

Therefore parabola properties are:

(h, k) = (5, - \frac{3}{2}), p = - \frac{3}{16}

Parabola is of the form

4 p (x - h) = (y - k)^{2}

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = - \frac{3}{2}

(y - 2)^{2} = 8 (x - 4)

y = (x + 5) (x + 4)

y = 8 x^{2} + 7 x - 1

y = x^{2} - 64

x = 3 y^{2}