Find the Asymptotes f(x)=(2x^2+6)/(2x^2+3x-2)
The problem asks for the determination of the asymptotes of the function f(x) = (2x^2 + 6) / (2x^2 + 3x - 2).
Asymptotes are lines that the graph of a function approaches but never actually reaches. There are two types of asymptotes typically considered for rational functions like the one provided:
Vertical Asymptotes occur at values of x where the denominator of the function is zero (assuming the numerator is not also zero at these points, which would potentially indicate a removable discontinuity instead).
Horizontal Asymptotes describe the behavior of the function as x approaches infinity or negative infinity. These are found by comparing the degrees of the numerator and denominator in the rational function and determining the end behavior of the function.
Additionally, there can be Oblique (or Slant) Asymptotes, which occur when the degree of the numerator is exactly one greater than the degree of the denominator, resulting in a non-horizontal, linear asymptote that the function approaches as x goes to infinity or negative infinity.
The question is essentially asking you to analyze the function and identify any of these asymptotes based on its algebraic form.
$f \left(\right. x \left.\right) = \frac{2 x^{2} + 6}{2 x^{2} + 3 x - 2}$
Answer
Solution:
Step 1:
Identify the values of $x$ that cause the function $\frac{2x^2 + 6}{2x^2 + 3x - 2}$ to be undefined, which are $x = -2$ and $x = \frac{1}{2}$.
Step 2:
Observe that as $x$ approaches $-2$ from the left, the function tends towards positive infinity, and from the right, it tends towards negative infinity. Hence, $x = -2$ is a vertical asymptote.
Step 3:
Similarly, as $x$ approaches $\frac{1}{2}$ from the left, the function tends towards negative infinity, and from the right, it tends towards positive infinity. Therefore, $x = \frac{1}{2}$ is also a vertical asymptote.
Step 4:
Compile a list of the vertical asymptotes, which are $x = -2$ and $x = \frac{1}{2}$.
Step 5:
Consider the general form of 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 horizontal asymptote depends on the relationship between $n$ and $m$:
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 6:
Determine the values of $n$ and $m$. For our function, $n = 2$ and $m = 2$.
Step 7:
Since $n$ equals $m$, the horizontal asymptote is found by the formula $y = \frac{a}{b}$. Given $a = 2$ and $b = 2$, the horizontal asymptote is $y = 1$.
Step 8:
There is no oblique asymptote since the degree of the numerator is not greater than the degree of the denominator.
Step 9:
Summarize all the asymptotes of the function:
- Vertical Asymptotes: $x = -2$, $x = \frac{1}{2}$
- Horizontal Asymptote: $y = 1$
- No Oblique Asymptotes
Knowledge Notes:
Asymptotes are lines that a graph of a function approaches as $x$ or $y$ goes to infinity. There are three types of asymptotes: vertical, horizontal, and oblique.
Vertical Asymptotes: Occur at values of $x$ where the function becomes undefined and the graph goes to positive or negative infinity. To find them, set the denominator equal to zero and solve for $x$.
Horizontal Asymptotes: These are horizontal lines that the graph approaches as $x$ goes to positive or negative infinity. The horizontal asymptote depends on the degrees of the polynomial in the numerator ($n$) and the denominator ($m$):
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 exactly one more than the degree of the denominator. The oblique asymptote can be found using polynomial long division.
In the given function $\frac{2x^2 + 6}{2x^2 + 3x - 2}$, the vertical asymptotes are found by setting the denominator to zero and solving for $x$. The horizontal asymptote is determined by comparing the degrees of the numerator and denominator, which are equal in this case, leading to a horizontal asymptote at $y = \frac{a}{b}$. Since the degrees are equal, there is no oblique asymptote.
