What is the axis y=-x^2+2x-8 ?

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axis y=-x^2+2x-8

axis

y = - x^{2} + 2 x - 8

x = 1

Solution steps

y = - x^{2} + 2 x - 8

Rewrite

y = - x^{2} + 2 x - 8

in the standard form

4 (- \frac{1}{4}) (y - (- 7)) = (x - 1)^{2}

Therefore parabola properties are:

(h, k) = (1, - 7), 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 = 1

y = - \frac{1}{2} (x + 3)^{2} - 1

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

y = + (3 - 2 x) (6 - x)

y^{2} - 16 x = 0

y = (x + 4)^{2} - 9