Find the Asymptotes f(x)=(x^2-8)/(2x^2-18)
The question is asking for the identification of the asymptotes of the function f(x) = (x^2 - 8)/(2x^2 - 18). Asymptotes are lines that the graph of the function approaches but does not touch as the magnitude of x increases infinitely in either the positive or negative direction. The question specifically requires the determination of both vertical and horizontal (and possibly oblique) asymptotes for the given rational function. Vertical asymptotes occur where the denominator of the function approaches zero, while horizontal or oblique asymptotes deal with the behavior of the function as x approaches infinity or negative infinity.
$f \left(\right. x \left.\right) = \frac{x^{2} - 8}{2 x^{2} - 18}$
Determine the values of $x$ for which $\frac{x^2 - 8}{2x^2 - 18}$ does not exist. These are $x = -3$ and $x = 3$.
Observe the behavior of $\frac{x^2 - 8}{2x^2 - 18}$ as $x$ approaches $-3$. As $x$ approaches $-3$ from the left, the expression tends towards positive infinity, and from the right, it tends towards negative infinity. Thus, $x = -3$ is a vertical asymptote.
Analyze the behavior of $\frac{x^2 - 8}{2x^2 - 18}$ as $x$ approaches $3$. As $x$ approaches $3$ from the left, the expression tends towards negative infinity, and from the right, it tends towards positive infinity. Hence, $x = 3$ is a vertical asymptote.
Compile a list of all vertical asymptotes: $x = -3$ and $x = 3$.
Review the conditions for horizontal asymptotes in a rational function $R(x) = \frac{ax^n}{bx^m}$, where $n$ is the degree of the numerator and $m$ is the degree of the denominator. The conditions are as follows:
If $n < m$, then the horizontal asymptote is $y = 0$.
If $n = m$, then the horizontal asymptote is $y = \frac{a}{b}$.
If $n > m$, there is no horizontal asymptote, but there may be an oblique asymptote.
Identify the degrees $n$ and $m$ of the numerator and denominator, respectively. Here, $n = 2$ and $m = 2$.
Since the degrees of the numerator and denominator are equal ($n = m$), the horizontal asymptote is given by $y = \frac{a}{b}$, where $a = 1$ and $b = 2$. Therefore, the horizontal asymptote is $y = \frac{1}{2}$.
Conclude that there are no oblique asymptotes since the degree of the numerator is not greater than the degree of the denominator.
Summarize the set of all asymptotes for the function:
Vertical Asymptotes: These occur at values of $x$ where the function becomes undefined, typically where the denominator of a rational function is zero.
Horizontal Asymptotes: These are horizontal lines that the graph of the function approaches as $x$ goes to positive or negative infinity. The rules for finding horizontal asymptotes are based on the degrees of the numerator ($n$) and denominator ($m$) of the rational function:
If $n < m$, the horizontal asymptote is $y = 0$.
If $n = m$, the horizontal asymptote is $y = \frac{a}{b}$, where $a$ and $b$ are the leading coefficients of the numerator and denominator, respectively.
If $n > m$, there is no horizontal asymptote.
Oblique Asymptotes: These occur when the degree of the numerator is one more than the degree of the denominator. The oblique asymptote can be found by performing polynomial long division.
To find the vertical asymptotes of a rational function, set the denominator equal to zero and solve for $x$.
To find the horizontal asymptote, compare the degrees of the numerator and denominator and apply the rules mentioned above.
It's important to note that the behavior of the function around vertical asymptotes can vary; it can approach infinity from both sides, negative infinity from both sides, or infinity from one side and negative infinity from the other.
In the given function $\frac{x^2 - 8}{2x^2 - 18}$, the vertical asymptotes are found by setting the denominator equal to zero, which gives $x = \pm 3$. The horizontal asymptote is determined by the fact that the degrees of the numerator and denominator are equal, resulting in $y = \frac{1}{2}$.