Over what interval is the function decreasing? f(x)=(3−x)^2 + 5

It decreases on the interval   (-infinity, 3)

This is the same as saying -infinity < x < 3, or simply x < 3

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Work Shown:

f(x) = (3−x)^2 + 5

f(x) = (-x+3)^2 + 5

f(x) = (-1(x-3))^2 + 5

f(x) = (-1)^2*(x-3)^2 + 5

f(x) = 1(x-3)^2 + 5

The last equation is in the form y = a(x-h)^2 + k

We have a = 1, h = 3 and k = 5

The vertex is (h,k) = (3,5)

Since 'a' is positive, this means the function decreases on the left and increases on the right (forming a parabola that opens upward). See below.

So the function is decreasing when x < 3, which represents everything to the left of the vertex.

We can expand that out to -infinity < x < 3 which converts to the interval notation (-infinity, 3)

the answer is y = 11.54

hope this : )