Final answer:
The graph of the function f(x) = x² + 3x + 5/x + 2 has a vertical asymptote at x = -2. However, the options provided for an oblique asymptote do not accurately represent the function as it is written because the degree of the numerator is not one higher than the denominator's degree.
Step-by-step explanation:
Understanding the Function f(x)
To find the correct graph of the function f(x) = x² + 3x + 5/x + 2, we need to consider both vertical and oblique asymptotes. Vertical asymptotes occur where the function is undefined, which in this case is when the denominator is zero. So, the vertical asymptote is at x = -2, since that would make the denominator 0. An oblique or slant asymptote can sometimes occur when the degree of the numerator is one higher than the denominator, and it's found by long division of the polynomials. However, the given function does not fit this exact scenario because the numerator and denominator are not such that the degree of the numerator is exactly one higher than the denominator.
Considering the provided options, all of them incorrectly assume an oblique asymptote, which suggests there might have been an error in forming the options or in the original function stated. Given the nature of the function, as x becomes very large or very negative, the x² term in the numerator will dominate, driving the value of the function upward, without approaching a straight line. However, if we ignore the fact that the form doesn't perfectly fit the case for an oblique asymptote and attempt long division, we would find that the leading term of the quotient (if we had a linear term in the numerator with a higher degree) would be x, indicating that if there were an oblique asymptote, it would resemble y = x + b, for some constant b.
Considering these analyses and disregarding the impossibility of an oblique asymptote for the function as written, the closest correct choice would be d. graph with vertical asymptote of ( x = -2 ), and oblique asymptote of ( y = x + 1 ).