What is the axis y=-8x^2+4x ?

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axis y=-8x^2+4x

axis

y = - 8 x^{2} + 4 x

x = \frac{1}{4}

Solution steps

y = - 8 x^{2} + 4 x

Rewrite

y = - 8 x^{2} + 4 x

in the standard form

4 (- \frac{1}{32}) (y - \frac{1}{2}) = (x - \frac{1}{4})^{2}

Therefore parabola properties are:

(h, k) = (\frac{1}{4}, \frac{1}{2}), p = - \frac{1}{32}

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{1}{4}

- 3 y = x^{2} - 8 x + 37

y = - 2 x^{2} + 12 x - 10

y = 2 x^{2} - 12 x + 18

x^{2} - 8 y = 0

y = - 9 x^{2} + 700 x - 6005