What's a function that has a vertical asymptote at x= -8 and x= 3 and a hole at (5,1/26)?

Answer 1

vertical asymptotes at x= -8 and x= 3
y = a / (x+8)(x-3)
a hole means that the function evaluates to 0/0 at x=5. So add that factor above and below
y = a(x-5) / (x+8)(x-3)(x-5)
Now, since y(5) = 1/26, we need to find a such that
a / (5+8)(5-3) = 1/26
clearly, a=1, so

y = (x-5) / (x+8)(x-3)(x-5)

does the trick.