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

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axis x>-(y+4)^2

axis

x > - (y + 4)^{2}

y = - 4

Solution steps

x > - (y + 4)^{2}

Rewrite

x > - (y + 4)^{2}

in the standard form:

4 (- \frac{1}{4}) (x - 0) = (y - (- 4))^{2}

Therefore parabola properties are:

(h, k) = (0, - 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^{2} + 2 x + 10

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

y = + \frac{17}{4} (x + 2)^{2}

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

x = (y - 4)^{2} + 3