Find the Asymptotes (x^2+2x-8)/(x^2-4)
The question is asking you to determine the lines that the graph of the function (x^2+2x-8)/(x^2-4) approaches as the values of x either become very large or very small, or as the function approaches undefined points. These lines are called asymptotes and can be vertical, horizontal, or oblique (slant). The goal is to identify these asymptotes from the given rational function by analyzing the behavior of the function at infinity and at points where the denominator equals zero.
$\frac{x^{2} + 2 x - 8}{x^{2} - 4}$
Answer
Solution:
Step 1:
Identify the values of $x$ for which the function $\frac{x^2 + 2x - 8}{x^2 - 4}$ does not exist. These are $x = -2$ and $x = 2$.
Step 2:
Examine the behavior of the function as $x$ approaches $-2$. As $x$ approaches $-2$ from the left, the function tends towards negative infinity, and from the right, it tends towards positive infinity. Thus, $x = -2$ is a vertical asymptote.
Step 3:
Consider a general rational function $f(x) = \frac{ax^n}{bx^m}$, where $n$ is the degree of the polynomial in the numerator and $m$ is the degree of the polynomial in the denominator. The rules for horizontal asymptotes are as follows:
If $n < m$, the horizontal asymptote is $y = 0$.
If $n = m$, the horizontal asymptote is $y = \frac{a}{b}$.
If $n > m$, there is no horizontal asymptote, but there may be an oblique asymptote.
Step 4:
Determine the degrees $n$ and $m$ for the given function. Here, $n = 2$ and $m = 2$.
Step 5:
Since the degrees of the numerator and denominator are equal ($n = m$), the horizontal asymptote can be found by dividing the leading coefficients. In this case, $a = 1$ and $b = 1$, so the horizontal asymptote is $y = \frac{1}{1} = 1$.
Step 6:
An oblique asymptote is not present because the degree of the numerator is not greater than the degree of the denominator.
Step 7:
Compile the list of asymptotes for the function:
- Vertical Asymptotes: $x = -2$
- Horizontal Asymptotes: $y = 1$
- No Oblique Asymptotes
Step 8:
(No further steps are required as all asymptotes have been identified.)
Solution:"The set of asymptotes for the function $\frac{x^2 + 2x - 8}{x^2 - 4}$ includes a vertical asymptote at $x = -2$ and a horizontal asymptote at $y = 1$. There are no oblique asymptotes."
Knowledge Notes:
Asymptotes are lines that the graph of a function approaches but never touches as the inputs or outputs grow without bound.
Vertical Asymptotes occur at values of $x$ where the function becomes undefined and the limits approach infinity or negative infinity.
Horizontal Asymptotes are found by comparing the degrees of the numerator and denominator polynomials in a rational function. If the degrees are equal, the horizontal asymptote is $y = \frac{a}{b}$, where $a$ and $b$ are the leading coefficients.
Oblique Asymptotes may occur when the degree of the numerator is one more than the degree of the denominator. In such cases, long division of the polynomials can be used to find the equation of the oblique asymptote.
Limits play a crucial role in determining the behavior of functions near their asymptotes, especially when evaluating whether a function tends towards infinity or negative infinity as it approaches a vertical asymptote.
