Question 909861
we have f(x) = (x+1)(x-3)(x-5)g(x) for some second degree polynomial g(x)
To determine g(x):
We need f(x) ≥ 0 for x ≤ 5.
==> (x + 1)(x - 3)(x - 5)g(x) ≥ 0 for x ≤ 5.
Since we need g(x) to be quadratic, we can take g(x) = A(x + 1)(x - 3) for some constant A and
Now, we have f(x) = A(x + 1)^2 (x - 3)^2 (x - 5) for some A. 
we use f(0) = A(1)^2 (-3)^2 (-5) = 15
15 = A*(1)*9*(-5)
15 = -45A
A = -1/3
therefore
f(x) = (-1/3)(x + 1)^2 (x - 3)^2 (x - 5)
this function can be graphed when multiplied out