Final answer:
The graph of y = (2x^3 - 16) / (2+1)(x^2 - 25) has potential vertical asymptotes at x = -1, x = 5, and x = -5, as these are the values that make the denominator zero. However, it is important to confirm each to ensure they are truly asymptotes by checking if the function approaches infinity near those points.
Step-by-step explanation:
The number of vertical asymptotes in the graph of the function y = (2x^3 - 16) / (2x^2 - 16) can be determined by looking at the values of x which make the denominator equal to zero, as these are the values where the function may approach infinity. The denominator is (2x^2 - 16), which can be factored as 2(x^2 - 8). Setting this equal to zero gives us x^2 - 8 = 0, or x^2 = 8. This implies that the graph has vertical asymptotes at x = sqrt(8) and x = -sqrt(8), which are approximately x = 2.828 and x = -2.828.
However, a typo in the original function was detected. The correct function as y = (2x^3 - 16) / ((x + 1)(x^2 - 25)) should be considered. The denominator factors to (x + 1)(x - 5)(x + 5), hence there are potential vertical asymptotes at x = -1, x = 5, and x = -5. To confirm if these values are indeed vertical asymptotes, you must verify that the function approaches infinity as x approaches each of these values.