Problem 27
Question
For the following exercises, describe the local and end behavior of the functions. $$ f(x)=\frac{-2 x}{x-6} $$
Step-by-Step Solution
Verified Answer
The function has a vertical asymptote at \( x = 6 \) and a horizontal asymptote at \( y = -2 \). As \( x \rightarrow \\pm\infty \), \( f(x) \rightarrow -2 \).
1Step 1: Identify Vertical Asymptotes
To find the vertical asymptotes, we need to set the denominator equal to zero and solve for \( x \). The function \( f(x) = \frac{-2x}{x-6} \) has a denominator of \( x-6 \). \( x-6 = 0 \) when \( x = 6 \). Hence, there is a vertical asymptote at \( x = 6 \).
2Step 2: Identify Horizontal Asymptotes
Evaluate the degrees of the numerator and denominator. Since both the numerator and denominator have the same degree (1), the horizontal asymptote is determined by the ratio of their leading coefficients. The leading coefficients are \(-2\) and \(1\), respectively, giving a horizontal asymptote of \( y = \frac{-2}{1} = -2 \).
3Step 3: Determine the Local Behavior Near Asymptotes
Check the behavior of \( f(x) \) as \( x \) approaches the vertical asymptote \( x = 6 \) from both sides. As \( x \rightarrow 6^+ \), the denominator \( x-6 \) approaches positive zero, so \( f(x) \rightarrow -fty \). As \( x \rightarrow 6^- \), \( x-6 \) approaches negative zero, so \( f(x) \rightarrow +fty \).
4Step 4: Determine End Behavior
Examine the limit of \( f(x) \) as \( x \rightarrow \pm\infty \). As \( x \rightarrow \infty \) or \( x \rightarrow -\infty \), the function approaches the horizontal asymptote: \( y = -2 \).
Key Concepts
Vertical AsymptotesHorizontal AsymptotesLocal BehaviorEnd Behavior
Vertical Asymptotes
Vertical asymptotes in rational functions occur where the denominator becomes zero because division by zero is undefined. In the function \( f(x) = \frac{-2x}{x-6} \), we identify the vertical asymptote by setting the denominator \( x-6 \) equal to zero. Solving for \( x \), we get \( x = 6 \). This means that as \( x \) approaches 6, the function's value heads towards infinity or negative infinity, depending on the direction of approach. This vertical line at \( x = 6 \) helps us understand the points where the function cannot exist and indicates a distinct change in behavior of the function.
Horizontal Asymptotes
Horizontal asymptotes tell us about the behavior of a function as \( x \) moves towards extremely large or small values. To find them in rational functions, compare the degrees of the numerator and the denominator. For \( f(x) = \frac{-2x}{x-6} \), both have the same degree of 1. When degrees are equal, the horizontal asymptote is determined by the ratio of the leading coefficients. Here, the leading coefficient of the numerator is \(-2\), and of the denominator is \(1\). Thus, the horizontal asymptote is \( y = \frac{-2}{1} = -2 \). As \( x \) nears infinity or negative infinity, \( f(x) \) gets close to \( -2 \). This horizontal line gives us insight into the function's limiting behavior at the far reaches of the x-axis.
Local Behavior
The local behavior of a rational function near its vertical asymptote can be quite dramatic. Consider the function \( f(x) = \frac{-2x}{x-6} \), which has a vertical asymptote at \( x = 6 \). To understand local behavior around this asymptote, observe how \( f(x) \) behaves as \( x \) approaches \( 6 \) from both sides:
- As \( x \rightarrow 6^+ \), \( x-6 \) becomes a very small positive number, making \( f(x) \) plunge towards negative infinity.
- As \( x \rightarrow 6^- \), \( x-6 \) becomes a very small negative number, causing \( f(x) \) to skyrocket towards positive infinity.
End Behavior
End behavior is how a function behaves as \( x \) increases or decreases without bound. For \( f(x) = \frac{-2x}{x-6} \), the end behavior is tied to its horizontal asymptote.
- As \( x \rightarrow \infty \), the function approaches the horizontal asymptote \( y = -2 \).
- The same occurs as \( x \rightarrow -\infty \).
Other exercises in this chapter
Problem 26
For the following exercises, use the vertex \((h, k)\) and a point on the graph \((x, y)\) to find the general form of the equation of the quadratic function. $
View solution Problem 27
For the following exercises, use the given information to find the unknown value. \(y\) varies directly as the square root of \(x\). When \(x=16,\) then \(y=4\)
View solution Problem 27
For the following exercises, use the Rational Zero Theorem to find all real zeros. $$ 2 x^{3}-3 x^{2}-32 x-15=0 $$
View solution Problem 27
For the following exercises, find the inverse of the functions. $$ f(x)=\frac{5 x+1}{2-5 x} $$
View solution