Find the Asymptotes f(x)=(e^(-2x))/(x-9)

Solution:

Step 1:

Identify the values for which $\frac{e^{-2x}}{x-9}$ does not exist. This occurs when $x=9$.

Step 2:

Observe that as $x$ grows large, $e^{-2x}$ tends towards zero, and since $x-9$ grows without bound, the fraction $\frac{e^{-2x}}{x-9}$ tends towards $0$.

Step 3:

Enumerate the horizontal asymptotes. The function approaches $y=0$ as $x$ goes to infinity.

Step 4:

Determine the presence of any slant asymptotes. Since the exponent of $e^{-2x}$ is not higher than that of $x-9$, there are no slant asymptotes.

Step 5:

Compile a list of all the asymptotes for the function:

Vertical Asymptotes: $x=9$ Horizontal Asymptotes: $y=0$ There are no slant asymptotes.

Knowledge Notes:

Asymptotes are lines that a graph approaches but never actually touches. They can be vertical, horizontal, or oblique (slant).

1. Vertical asymptotes occur where the function is undefined, typically where the denominator of a fraction is zero.

2. Horizontal asymptotes are found by examining the behavior of a function as $x$ approaches infinity or negative infinity. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y=0$.

3. Oblique (slant) asymptotes occur when the degree of the numerator is exactly one greater than the degree of the denominator. To find an oblique asymptote, you would typically perform polynomial long division.

In the given function $\frac{e^{-2x}}{x-9}$, the vertical asymptote is at $x=9$, where the denominator is zero. The horizontal asymptote is $y=0$, as the numerator approaches zero faster than the denominator approaches infinity. There is no oblique asymptote because the numerator is an exponential function, not a polynomial, and its degree is not one greater than that of the polynomial denominator.