What is the axis y=x^2-x-12 ?

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axis y=x^2-x-12

axis

y = x^{2} - x - 12

x = \frac{1}{2}

Solution steps

y = x^{2} - x - 12

Rewrite

y = x^{2} - x - 12

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{49}{4})) = (x - \frac{1}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{1}{2}, - \frac{49}{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 = (3 y - 1)^{2}

y = \frac{( - 7 x + 3 ) ( 9 x + 4 )}{24}

0 = (y - 1)^{2} - 16 (x + 4)

y = - 1 + 3 x + x^{2}

y = 7 x^{2} - 5 x + 9