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

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axis y=+14-x^2

axis

y = + 14 - x^{2}

x = 0

Solution steps

y = 14 - x^{2}

Rewrite

y = 14 - x^{2}

in the standard form:

4 (- \frac{1}{4}) (y - 14) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 14), p = - \frac{1}{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 = 0

y = 2 x^{2} + 28 x + 105

y = 4 \cdot x^{2} - 3 \cdot x - 12

x \leq y^{2}

y = (2 X + 5) (3 X - 7)

f (x) = - x^{2} - 4 x + 5