Final answer:
The function f(x) = (3x² +36x+81)/(x² -x-12) has a horizontal asymptote at y = 3, and vertical asymptotes at x = 4 and x = -3.
Step-by-step explanation:
Finding Practical Asymptotes of a Function. To find the practical asymptotes of the function f(x) = (3x² +36x+81)/(x² -x-12), we need to look at the behavior of the function as x approaches certain critical values. For horizontal asymptotes, we compare the degrees of the polynomials in the numerator and denominator. Since the degrees are the same (both are second degree polynomials), the horizontal asymptote will be the ratio of the leading coefficients, which is 3/1, giving us a horizontal asymptote at y = 3. Vertical asymptotes occur where the denominator equals zero and the function isn't defined. Factoring the denominator x² - x - 12, we get (x - 4)(x + 3). Setting each factor to zero gives us possible vertical asymptotes at x = 4 and x = -3. However, there are no common factors in the numerator and denominator, which means these are indeed the vertical asymptotes, as no simplification can remove these points of discontinuity. Thus, the function has vertical asymptotes at x = 4 and x = -3, and a horizontal asymptote at y = 3.