What is the axis y=-x^2+8x-7 ?

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axis y=-x^2+8x-7

axis

y = - x^{2} + 8 x - 7

x = 4

Solution steps

y = - x^{2} + 8 x - 7

Rewrite

y = - x^{2} + 8 x - 7

in the standard form

4 (- \frac{1}{4}) (y - 9) = (x - 4)^{2}

Therefore parabola properties are:

(h, k) = (4, 9), 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 = 4

y = x^{2} - 6 x + 9

y = - (x - 4)^{2} - 24

y = - 0.05 x^{2} + 10

y^{2} + 24 y - 9 x + 225 = 0

(3 x - 2)^{2} = 84 y - 112