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

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

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

axis

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

y = 1

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

(h, k) = (1, 1), p = - 4

Parabola is of the form

4 p (x - h) = (y - k)^{2}

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = 1

y = \frac{- 7 + 2 x ^{2} - 4 x}{3}

y = x^{2} - 10 x - 7

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

y = x^{2} - 6 x + 18

x + 5 + (y - 3)^{2} = 0