What is the axis y=-16x^2+48x+100 ?

The axis y=-16x^2+48x+100 is x= 3/2

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axis y=-16x^2+48x+100

axis

y = - 16 x^{2} + 48 x + 100

x = \frac{3}{2}

Solution steps

y = - 16 x^{2} + 48 x + 100

Rewrite

y = - 16 x^{2} + 48 x + 100

in the standard form

4 (- \frac{1}{64}) (y - 136) = (x - \frac{3}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{3}{2}, 136), 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 = \frac{3}{2}

y^{2} - 10 x + 4 y - 16 = 0

y = - 12 x^{2} + 156 x

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

y = 2 x^{2} - x

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