What is the axis y=-2(x-4)^2+5 ?

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axis y=-2(x-4)^2+5

axis

y = - 2 (x - 4)^{2} + 5

x = 4

Solution steps

y = - 2 (x - 4)^{2} + 5

Rewrite

y = - 2 (x - 4)^{2} + 5

in the standard form:

4 (- \frac{1}{8}) (y - 5) = (x - 4)^{2}

Therefore parabola properties are:

(h, k) = (4, 5), 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 = 4

x = (y + 1) (y + 2)

4 x^{2} - 128 y = 0

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

y = (x - 13)^{2}

y = x^{2} - 5 x - 84