What is the axis (x+2)^2=-8(y+3) ?

The axis (x+2)^2=-8(y+3) is x=-2

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axis (x+2)^2=-8(y+3)

axis

(x + 2)^{2} = - 8 (y + 3)

x = - 2

Solution steps

(x + 2)^{2} = - 8 (y + 3)

Rewrite

(x + 2)^{2} = - 8 (y + 3)

in the standard form:

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

Therefore parabola properties are:

(h, k) = (- 2, - 3), 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 = - 2

y = \frac{1 + x ^{2} - 8 x}{5}

y^{2} + 6 x - 4 y - 11 = 0

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

+ 3 y^{2} = 24 x

y = - 2 x^{2} - 8 x - 3