Question

Horizontal asymptotes Determine \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x)\) for the following functions. Then give the horizontal asymptotes of \(f(\text {if any})\). $$f(x)=\frac{6 x^{2}-9 x+8}{3 x^{2}+2}$$

Short answer

Answer: The horizontal asymptote of the function is \(y=2\).

Step-by-step solution

Identify the degree of the numerator and the denominator

In this case, we have a rational function where the numerator has degree 2 and the denominator also has degree 2.

Divide the numerator and denominator by the highest power of x

Since the highest power of x is \(x^2\), we will divide both numerator and denominator by \(x^2\). \(f(x) = \frac{6x^2 - 9x + 8}{3x^2 + 2} = \frac{\frac{6x^2}{x^2} - \frac{9x}{x^2} + \frac{8}{x^2}}{\frac{3x^2}{x^2} + \frac{2}{x^2}} = \frac{6 - \frac{9}{x} + \frac{8}{x^2}}{3 + \frac{2}{x^2}}\)

Find the limits as x approaches \(\infty\) and \(-\infty\)

When x is a very large positive or negative number, the terms where x is in the denominator become very small and tend to 0. So we have: \(\lim_{x\rightarrow \infty}\frac{6 - \frac{9}{x} + \frac{8}{x^2}}{3 + \frac{2}{x^2}}=\frac{6 - 0 + 0}{3 + 0}=\frac{6}{3}=2\) \(\lim_{x\rightarrow -\infty}\frac{6 - \frac{9}{x} + \frac{8}{x^2}}{3 + \frac{2}{x^2}}=\frac{6 - 0 + 0}{3 + 0}=\frac{6}{3}=2\)

Identify horizontal asymptotes

Since the limits are the same as x approaches \(\infty\) and \(-\infty\), we have a horizontal asymptote at \(y=2\). So the horizontal asymptote of the function \(f(x) = \frac{6x^2 - 9x + 8}{3x^2 + 2}\) is \(y=2\).

Key concepts

Limits of Functions

Understanding the concept of limits is crucial when analyzing the behavior of functions, especially as they move toward infinity or negative infinity. Limits are used to describe the value that a function approaches as the input (or 'x' value) gets infinitely close to a certain point. Mathematically, if we say \( \lim_{x\to a}f(x) = L \) it means as 'x' gets closer and closer to 'a', the function 'f(x)' approaches the value 'L'.

In the context of the provided exercise, limits help us determine the horizontal asymptotes of a rational function. When the degrees of the numerator and denominator are equal as in our function \(f(x) = \frac{6x^2 - 9x + 8}{3x^2 + 2}\), we find the limits by simplifying the function and observing the values that remain once the terms with 'x' in the denominator are disregarded as they approach zero.

Rational Functions

A rational function is a fraction composed of two polynomials, where one polynomial is in the numerator and another is in the denominator. The general form is \(f(x) = \frac{p(x)}{q(x)}\), where both \(p(x)\) and \(q(x)\) are polynomials. The behavior of these functions is particularly interesting when 'x' grows larger in either direction – positively or negatively.

In our example \(f(x) = \frac{6x^2 - 9x + 8}{3x^2 + 2}\), we deal with a rational function where the polynomials in the numerator and denominator are both second-degree polynomials. The degree of the polynomials present factors into how we look for horizontal asymptotes, indicating that in cases like these, the horizontal asymptote will depend on the leading coefficients of the polynomials.

Asymptotic Behavior

The asymptotic behavior of a function involves the way the function acts near a given line, called an asymptote, as the inputs get very large or very small. Horizontal asymptotes are specific types that occur when the value of the function approaches a constant value as 'x' goes to infinity or negative infinity. These asymptotes can be thought of as the 'behavioral boundaries' of a function on a graph.

In the exercise provided, the horizontal asymptote of the function \(f(x)\) is determined by calculating the limits as 'x' approaches infinity and negative infinity. The calculations show that as 'x' becomes very large in either direction, \(f(x)\) approaches the value of 2. Therefore, the line \(y=2\) is the horizontal asymptote, which serves as a 'flat line' that the curve of the graph gets closer and closer to but never actually touches.