What is the axis y=-x^2+120x+1300 ?

The axis y=-x^2+120x+1300 is x=60

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axis y=-x^2+120x+1300

axis

y = - x^{2} + 120 x + 1300

x = 60

Solution steps

y = - x^{2} + 120 x + 1300

Rewrite

y = - x^{2} + 120 x + 1300

in the standard form

4 (- \frac{1}{4}) (y - 4900) = (x - 60)^{2}

Therefore parabola properties are:

(h, k) = (60, 4900), 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 = 60

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

y = 0.5 (x + 1)^{2} - 2

y = x^{2} - 26 x + 65

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

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