What is the axis x=(4y+3)(2y-5) ?

The axis x=(4y+3)(2y-5) is y= 7/8

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axis x=(4y+3)(2y-5)

axis

x = (4 y + 3) (2 y - 5)

y = \frac{7}{8}

Solution steps

x = (4 y + 3) (2 y - 5)

Rewrite

x = (4 y + 3) (2 y - 5)

in the standard form

4 \cdot \frac{1}{32} (x - (- \frac{169}{8})) = (y - \frac{7}{8})^{2}

Therefore parabola properties are:

(h, k) = (- \frac{169}{8}, \frac{7}{8}), p = \frac{1}{32}

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{7}{8}

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

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

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

y = + (3 - 2 x) (6 - x)

y^{2} - 16 x = 0