What is the axis (x-1)^2=-8(y-4) ?

The axis (x-1)^2=-8(y-4) is x=1

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axis (x-1)^2=-8(y-4)

axis

(x - 1)^{2} = - 8 (y - 4)

x = 1

Solution steps

(x - 1)^{2} = - 8 (y - 4)

Rewrite

(x - 1)^{2} = - 8 (y - 4)

in the standard form:

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

Therefore parabola properties are:

(h, k) = (1, 4), p = - 2

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

(2 x - 7)^{2} + 8

x = 2 y^{2} - 1 - (5 y - 6)

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

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

y = \frac{x ^{2} - 4 x}{11}