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

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axis y=+1-x-x^2

axis

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

x = - \frac{1}{2}

Solution steps

y = 1 - x - x^{2}

Rewrite

y = 1 - x - x^{2}

in the standard form

4 (- \frac{1}{4}) (y - \frac{5}{4}) = (x - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{2}, \frac{5}{4}), 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 = - \frac{1}{2}

x = \frac{y ^{2}}{8}

y = 18 x^{2} - 14 x + 9

y = - (- 12 x + 13)^{2} - 1

y = 3 x^{2} - 5

y = 3 x^{2} - 1