Problem 480
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 local behavior shows a vertical asymptote at \(x=6\) with \(f(x)\to +\infty\) as \(x\to 6^+\) and \(f(x)\to -\infty\) as \(x\to 6^-\). The end behavior shows \(f(x)\to 2\) as \(x\to \pm\infty\).
1Step 1: Identify Vertical Asymptotes
The function \(f(x) = \frac{2x}{x-6}\) can potentially have vertical asymptotes where the denominator is zero. Set the denominator equal to zero and solve for \(x\):\[ x - 6 = 0 \]Solving gives \(x = 6\). Therefore, there is a vertical asymptote at \(x = 6\).
2Step 2: Determine Horizontal Asymptotes
For rational functions of the form \(\frac{p(x)}{q(x)}\), where \(p(x)\) and \(q(x)\) are polynomials, the horizontal asymptote depends on the degrees of these polynomials. Here, both the numerator and the denominator have degree 1 (linear polynomials). The horizontal asymptote is found by dividing the leading coefficients:\[ y = \frac{2}{1} = 2 \]So, \(y = 2\) is the horizontal asymptote.
3Step 3: Analyze Local Behavior Near Asymptotes
Consider the behavior of \(f(x)\) as \(x\) approaches the vertical asymptote at \(x=6\) from both the left and right sides.- As \(x \to 6^+\), \(f(x)\) approaches \(+\infty\) since the function moves from slightly larger than 6.- As \(x \to 6^-\), \(f(x)\) approaches \(-\infty\) as the function comes from slightly smaller values than 6.
4Step 4: Determine End Behavior for Very Large or Small x
Analyze \(f(x)\) as \(x\) approaches infinity (large positive numbers) and negative infinity (large negative numbers):- As \(x \to \infty\) or \(x \to -\infty\), the function \(f(x)\) approaches the horizontal asymptote \(y=2\).Thus, the end behavior of the function is that it approaches \(y = 2\) as \(x\) goes to plus or minus infinity.
Key Concepts
Vertical AsymptotesHorizontal AsymptotesEnd Behavior
Vertical Asymptotes
In the world of rational functions, vertical asymptotes are peculiar lines where the function shoots up to positive or negative infinity. Think of them as invisible barriers that the graph of the function cannot cross. Identifying these vertical asymptotes involves spotting the x-values that make the denominator zero because division by zero is undefined in mathematics.
Consider the function \(f(x) = \frac{2x}{x-6}\). To find a vertical asymptote, set the denominator, \(x - 6\), to zero. Solving the equation \(x - 6 = 0\) gives \(x = 6\). Thus, there's a vertical asymptote at \(x = 6\).
The behavior around this asymptote tells us the following:
Consider the function \(f(x) = \frac{2x}{x-6}\). To find a vertical asymptote, set the denominator, \(x - 6\), to zero. Solving the equation \(x - 6 = 0\) gives \(x = 6\). Thus, there's a vertical asymptote at \(x = 6\).
The behavior around this asymptote tells us the following:
- As \(x\) approaches 6 from the left (just a bit smaller than 6), the function values plunge to negative infinity.
- As \(x\) approaches 6 from the right (just a bit larger than 6), the values climb toward positive infinity.
Horizontal Asymptotes
Horizontal asymptotes provide a peek into how a function behaves at its extremes—the grand scale of things as \(x\) strides infinitely in either the positive or negative direction. For rational functions, these horizontal lines show the values that the function approaches but never actually touches at infinity.
For the rational function \(f(x) = \frac{2x}{x-6}\), determining the horizontal asymptote involves comparing the degrees of the numerator and the denominator. Both have a degree of 1 here, so we consider the leading coefficients, which are 2 and 1, respectively.
When the degrees are equal, the horizontal asymptote is simply the ratio of these leading coefficients: \(y = \frac{2}{1} = 2\).
Therefore, \(y = 2\) is the horizontal asymptote, indicating that as \(x\) becomes very large in the positive or negative sense, the function values hover around 2 without really reaching it. A horizontal asymptote gives us a stability zone, showing where the function tends to stabilize.
For the rational function \(f(x) = \frac{2x}{x-6}\), determining the horizontal asymptote involves comparing the degrees of the numerator and the denominator. Both have a degree of 1 here, so we consider the leading coefficients, which are 2 and 1, respectively.
When the degrees are equal, the horizontal asymptote is simply the ratio of these leading coefficients: \(y = \frac{2}{1} = 2\).
Therefore, \(y = 2\) is the horizontal asymptote, indicating that as \(x\) becomes very large in the positive or negative sense, the function values hover around 2 without really reaching it. A horizontal asymptote gives us a stability zone, showing where the function tends to stabilize.
End Behavior
End behavior describes the destiny of a function's journey as it stretches towards infinity or minus infinity. It's crucial for graphing because it hints at what happens as \(x\) becomes very large or very small.
For our function \(f(x) = \frac{2x}{x-6}\), let's explore this horizon.
For our function \(f(x) = \frac{2x}{x-6}\), let's explore this horizon.
- As \(x\) heads toward positive infinity, \(f(x)\) sneaks closer to the value of the horizontal asymptote, which is \(y = 2\).
- Similarly, as \(x\) drifts into negative infinity, the function also dwindles towards the same asymptote, \(y = 2\).
Other exercises in this chapter
Problem 478
For the following exercises, find the \(x\) - and \(y\) -intercepts for the functions. $$f(x)=\frac{94-2 x^{2}}{3 x^{2}-12}$$
View solution Problem 479
For the following exercises, describe the local and end behavior of the functions. $$f(x)=\frac{x}{2 x+1}$$
View solution Problem 481
For the following exercises, describe the local and end behavior of the functions. $$f(x)=\frac{-2 x}{x-6}$$
View solution Problem 482
For the following exercises, describe the local and end behavior of the functions. $$f(x)=\frac{x^{2}-4 x+3}{x^{2}-4 x-5}$$
View solution