What is the axis y=-1/2 (x+3)^2-1 ?

The axis y=-1/2 (x+3)^2-1 is x=-3

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axis y=-1/2 (x+3)^2-1

axis

y = - \frac{1}{2} (x + 3)^{2} - 1

x = - 3

Solution steps

y = - \frac{1}{2} (x + 3)^{2} - 1

Rewrite

y = - \frac{1}{2} (x + 3)^{2} - 1

in the standard form:

4 (- \frac{1}{2}) (y - (- 1)) = (x - (- 3))^{2}

Therefore parabola properties are:

(h, k) = (- 3, - 1), 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 = - 3

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

y = + (3 - 2 x) (6 - x)

y^{2} - 16 x = 0

y = (x + 4)^{2} - 9

x = - \frac{( y + 3 ) ^{2}}{20} + 1