Final answer:
The vertical asymptotes of the function f(x) = (x² + 1)/(8x - 2x²) occur where the denominator is zero but the numerator is not, which are at x = 0 and x = 4.
Step-by-step explanation:
The vertical asymptotes of a function f(x) are found by determining the values of x that make the denominator equal to zero, as long as they do not also make the numerator zero at the same values because such points would be holes rather than asymptotes.
In the function f(x) = (x² + 1)/(8x - 2x²), the denominator is 8x - 2x². Setting the denominator to zero gives us the equation 8x - 2x² = 0. Factoring out the common term 2x results in 2x(4 - x) = 0, which gives us two values for x: x = 0 and x = 4.
However, we must also check the numerator x² + 1, which is never zero since the smallest value x² can have is zero, making the smallest value of x² + 1 equal to one. Therefore, the vertical asymptotes of this function are at x = 0 and x = 4, because only the denominator is zero at these points, and the numerator is not.