What is the axis x=4(y+6)+3(2+y+3y^2)-2y^2 ?

The axis x=4(y+6)+3(2+y+3y^2)-2y^2 is y=-1/2

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axis x=4(y+6)+3(2+y+3y^2)-2y^2

axis

x = 4 (y + 6) + 3 (2 + y + 3 y^{2}) - 2 y^{2}

y = - \frac{1}{2}

Solution steps

x = 4 (y + 6) + 3 (2 + y + 3 y^{2}) - 2 y^{2}

Rewrite

x = 4 (y + 6) + 3 (2 + y + 3 y^{2}) - 2 y^{2}

in the standard form

4 \cdot \frac{1}{28} (x - \frac{113}{4}) = (y - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (\frac{113}{4}, - \frac{1}{2}), p = \frac{1}{28}

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{1}{2}

y = (x + 1) (2 x + 5)

y = \frac{1}{4} x^{2} - 3 x - 7

y = 49 x^{2} - 64

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

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

What is the axis x=4(y+6)+3(2+y+3y^2)-2y^2 ?
The axis x=4(y+6)+3(2+y+3y^2)-2y^2 is y=-1/2