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

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

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

axis

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

x = - 1

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

(h, k) = (- 1, - 2), p = \frac{1}{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

y = x^{2} + 81

(y - 1)^{2} = - 16 (x - 1)

y = \frac{- 7 + 2 x ^{2} - 4 x}{3}

y = x^{2} - 10 x - 7

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