What is the axis 4x^2-12x-16y+41=0 ?

The axis 4x^2-12x-16y+41=0 is x= 3/2

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axis 4x^2-12x-16y+41=0

axis

4 x^{2} - 12 x - 16 y + 41 = 0

x = \frac{3}{2}

Solution steps

4 x^{2} - 12 x - 16 y + 41 = 0

Rewrite

4 x^{2} - 12 x - 16 y + 41 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (\frac{3}{2}, 2), p = 1

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 = (y - 12)^{2} - 156

x = \frac{1}{4} y^{2}

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

\frac{y - 2}{4} - \frac{x ^{2}}{9} = 1

y = 2 x^{2} + 24 x - 6