What is the axis y=-2(x-3)^2+2 ?

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axis y=-2(x-3)^2+2

axis

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

x = 3

Solution steps

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

Rewrite

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

in the standard form:

4 (- \frac{1}{8}) (y - 2) = (x - 3)^{2}

Therefore parabola properties are:

(h, k) = (3, 2), p = - \frac{1}{8}

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 = 3

y^{2} + 4 x = 0

y = - 9 x^{2} - 7

y = 3 x^{2} - 35 x

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

(y - 3)^{2} = - 20 (x + 2)