What is the axis (x+3)^2=4(y-1) ?

The axis (x+3)^2=4(y-1) is x=-3

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axis (x+3)^2=4(y-1)

axis

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

x = - 3

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

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

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

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

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

y = 2 x^{2} - 7

(x + 1)^{2} = 12 y

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