Final answer:
The function f(x) has a horizontal asymptote at y = 6 and a vertical asymptote at x = 3. The function h(x) does not have traditional horizontal or vertical asymptotes as it is dominated by exponential growth and decay.
Step-by-step explanation:
To find the horizontal and vertical asymptotes of the provided functions, we must examine the behavior of the functions as x approaches infinity and the points where the functions are undefined due to denominator being zero.
f(x) = \( \frac{12x}{2x - 6} \)
For horizontal asymptotes, we look at the limits as x approaches infinity. Since the degrees of the numerator and denominator are the same, the horizontal asymptote can be found by dividing the leading coefficients. As x approaches infinity:
\( \lim_{x \to \infty} \frac{12x}{2x - 6} = \frac{12}{2} = 6 \)
Thus, the horizontal asymptote is y = 6.
For vertical asymptotes, they occur where the function is undefined:
2x - 6 = 0
x = 3
As x approaches 3, the denominator approaches 0, and the function approaches infinity. Therefore, the vertical asymptote is x = 3.
h(x) = \( e^{\frac{x}{3}} - e^{2x} \)
For the function h(x), it is not rational, so it does not have vertical asymptotes in the same way as rational functions. However, we can still investigate the limits. As x approaches infinity, \( e^{\frac{x}{3}} \) and \( e^{2x} \) both grow towards infinity, but \( e^{2x} \) grows much faster. Hence, the function h(x) will be dominated by \( -e^{2x} \), and:
\( \lim_{x \to \infty} (e^{\frac{x}{3}} - e^{2x}) = -\infty \)
Thus, h(x) will not possess a horizontal asymptote as it approaches negative infinity. As x approaches negative infinity, \( e^{\frac{x}{3}} \) approaches 0 faster than \( -e^{2x} \) approaches 0. The limit of h(x) as x approaches negative infinity is therefore 0, but since the function becomes negative before crossing zero, there's no horizontal asymptote.