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

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

axis

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

y = 4

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

(h, k) = (1, 4), p = - \frac{1}{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 = 4

y = (x + 1) (x + 2)

y = x^{2} + x - 12

x = y^{2} - 2

y = x^{2} - 8 x + 7

(x - 5)^{2} = - 70 (y - 1)