What is the axis x^2=-14(y+1) ?

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axis x^2=-14(y+1)

axis

x^{2} = - 14 (y + 1)

x = 0

Solution steps

x^{2} = - 14 (y + 1)

Rewrite

x^{2} = - 14 (y + 1)

in the standard form:

4 (- \frac{7}{2}) (y - (- 1)) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, - 1), p = - \frac{7}{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 = 0

x = - 8 y^{2}

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

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

y = 4 x^{2} + 2

y = x^{2} + 4 x