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

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axis y=-x^2+8x

axis

y = - x^{2} + 8 x

x = 4

Solution steps

y = - x^{2} + 8 x

Rewrite

y = - x^{2} + 8 x

in the standard form

4 (- \frac{1}{4}) (y - 16) = (x - 4)^{2}

Therefore parabola properties are:

(h, k) = (4, 16), p = - \frac{1}{4}

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 = 4

y = - 16 x^{2} + 40 x + 2

x = - 2 y + y (y - 2) - 2 y (y - 1) - (y - 3)

y = - 0.5 x^{2} + 5.5

y = \frac{1}{2} x^{2} + 9 x + 8

x = (y - 7)^{2} + 1