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

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.

To find a function with the given characteristics, we need to understand the properties of vertical asymptotes and holes.

1. Vertical Asymptotes: A vertical asymptote is a vertical line where the function approaches infinity or negative infinity as the input approaches a specific value. In this case, we have vertical asymptotes at x = -8 and x = 3.

2. Holes: A hole (also known as a removable discontinuity) occurs when a function has a removable point of discontinuity. In this case, we have a hole at the point (5, 1/26).

To construct a function with these characteristics, we can follow these steps:

Step 1: Identify the factors of the function that cause the vertical asymptotes.
Given the vertical asymptotes at x = -8 and x = 3, we can write the factors as (x + 8) and (x - 3).

Step 2: Determine the factor that corresponds to the hole.
We know that there is a hole at (5, 1/26). To create this hole, we need a factor of (x - 5) and its corresponding denominator factor in the function.

Step 3: Combine the factors to form the function.
We multiply all the factors together to get the complete function.

The resulting function is:

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

This function has vertical asymptotes at x = -8 and x = 3, and a hole at (5, 1/26).

Please note that this function is only an example, and there could be multiple functions that satisfy the given characteristics.