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

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axis y=x^2+x-9

axis

y = x^{2} + x - 9

x = - \frac{1}{2}

Solution steps

y = x^{2} + x - 9

Rewrite

y = x^{2} + x - 9

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{37}{4})) = (x - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{2}, - \frac{37}{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}

2 y^{2} - x - 12 y + 18 = 0

y = (x + 6) (x - 2)

x = (y + 2) (y - 3)

x^{2} - 6 x + 5 y = 34

(x + 3)^{2} = 8 (y - 5)