What is the axis (x+3)^2=8(y-5) ?

The axis (x+3)^2=8(y-5) is x=-3

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axis (x+3)^2=8(y-5)

axis

(x + 3)^{2} = 8 (y - 5)

x = - 3

Solution steps

(x + 3)^{2} = 8 (y - 5)

Rewrite

(x + 3)^{2} = 8 (y - 5)

in the standard form:

4 \cdot 2 (y - 5) = (x - (- 3))^{2}

Therefore parabola properties are:

(h, k) = (- 3, 5), 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 = - 3

y = 4 x^{2} - 12 x + 10

y = x^{2} + 3 x + 2

y = (2 x + 5) (x + 6)

y = x^{2} + 3 x + 1

y = x^{2} + x + 1