What is the axis y=160x-16x^2 ?

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axis y=160x-16x^2

axis

y = 160 x - 16 x^{2}

x = 5

Solution steps

y = 160 x - 16 x^{2}

Rewrite

y = 160 x - 16 x^{2}

in the standard form

4 (- \frac{1}{64}) (y - 400) = (x - 5)^{2}

Therefore parabola properties are:

(h, k) = (5, 400), p = - \frac{1}{64}

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 = 5

x^{2} + 6 x + 3 y + 2 = 0

y = - 2 x^{2} - 3 x + 5

y = - 0.1 x^{2} + 0.8 x + 98.8

y = x^{2} - 4 x - 5

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