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

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axis y=-x^2+2x+10

axis

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

x = 1

Solution steps

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

Rewrite

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

in the standard form

4 (- \frac{1}{4}) (y - 11) = (x - 1)^{2}

Therefore parabola properties are:

(h, k) = (1, 11), 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 = 5 x^{2} + x - 4

y = + \frac{17}{4} (x + 2)^{2}

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

x = (y - 4)^{2} + 3

y = 2.1 + 4.4 x - 0.7 x^{2}