What is the axis y=-1/4 x^2+8x-4 ?

The axis y=-1/4 x^2+8x-4 is x=16

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axis y=-1/4 x^2+8x-4

axis

y = - \frac{1}{4} x^{2} + 8 x - 4

x = 16

Solution steps

y = - \frac{1}{4} x^{2} + 8 x - 4

Rewrite

y = - \frac{1}{4} x^{2} + 8 x - 4

in the standard form

4 (- 1) (y - 60) = (x - 16)^{2}

Therefore parabola properties are:

(h, k) = (16, 60), p = - 1

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 = 16

y = x^{2} - 10 x + 23

y^{2} + 8 x - 4 y + 20 = 0

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

y = 3 x^{2} + 6

y^{2} - 6 y - 4 x + 5 = 0