What is the axis y=-1/16 (x-1)^2-1 ?

The axis y=-1/16 (x-1)^2-1 is x=1

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axis y=-1/16 (x-1)^2-1

axis

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

x = 1

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

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

y = x^{2} + 7 x + 30

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

y = 6 x - x^{2} - 5

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

y = x^{2} + 2 x - 1