What is the axis x=+(y+1)^2-9 ?

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axis x=+(y+1)^2-9

axis

x = + (y + 1)^{2} - 9

y = - 1

Solution steps

x = (y + 1)^{2} - 9

Rewrite

x = (y + 1)^{2} - 9

in the standard form:

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

Therefore parabola properties are:

(h, k) = (- 9, - 1), p = \frac{1}{4}

Parabola is of the form

4 p (x - h) = (y - k)^{2}

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = - 1

y = x^{2} - 3 x - 10

4 x = y^{2} - 4 y

3 x^{2} + 19 y = 0

y = 3 x^{2} + 6 x + 10

F (x) = 2 x^{2} + 2 x + 3