What is the axis (x+3)^2=16(y-9) ?

The axis (x+3)^2=16(y-9) is x=-3

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axis (x+3)^2=16(y-9)

axis

(x + 3)^{2} = 16 (y - 9)

x = - 3

Solution steps

(x + 3)^{2} = 16 (y - 9)

Rewrite

(x + 3)^{2} = 16 (y - 9)

in the standard form:

4 \cdot 4 (y - 9) = (x - (- 3))^{2}

Therefore parabola properties are:

(h, k) = (- 3, 9), p = 4

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 = - 0.1 x^{2} + 1.4 x + 98.4

y = 4 x^{2} + 2

y = x^{2} + 4 x

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

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