What is the axis y=x^2+10x-11 ?

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axis y=x^2+10x-11

axis

y = x^{2} + 10 x - 11

x = - 5

Solution steps

y = x^{2} + 10 x - 11

Rewrite

y = x^{2} + 10 x - 11

in the standard form

4 \cdot \frac{1}{4} (y - (- 36)) = (x - (- 5))^{2}

Therefore parabola properties are:

(h, k) = (- 5, - 36), 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 = - 5

x = (y + 4)^{2}

y = + x^{2} - 9

y = + \frac{9 x ^{2} - 6 x + 3}{3} x

x^{2} - 12 x + 16 y + 68 = 0

y = 2 x^{2} - 4 x