What is the axis (x+9)^2+2(y-4)=0 ?

The axis (x+9)^2+2(y-4)=0 is x=-9

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axis (x+9)^2+2(y-4)=0

axis

(x + 9)^{2} + 2 (y - 4) = 0

x = - 9

Solution steps

(x + 9)^{2} + 2 (y - 4) = 0

Rewrite

(x + 9)^{2} + 2 (y - 4) = 0

in the standard form:

4 (- \frac{1}{2}) (y - 4) = (x - (- 9))^{2}

Therefore parabola properties are:

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

x^{2} = 36 y

y = - 0.0000848 x^{2} - 0.006005 x + 1.004

y = 2 x^{2} - 5 x + 1

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

y = + x^{2} + 4 x + 7