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

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

axis

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

y = 2

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

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

y = 3 (x - 1) (x + 2)

x^{2} + 8 x + 12

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

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

- x^{2} + 10 x - 24