Chapter 10: Problem 43
Find all asymptotes, \(x\) -intercepts, and \(y\) -intercepts for the graph of each rational function and sketch the graph of the function. $$f(x)=\frac{x+1}{x^{2}}$$
Short Answer
Expert verified
Vertical asymptote: x=0. Horizontal asymptote: y=0. x-intercept: (-1, 0). No y-intercept.
Step by step solution
01
- Find Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero, and the numerator is non-zero. For the function \(f(x)=\frac{x+1}{x^{2}}\), set the denominator equal to zero: \[ x^2 = 0 \]Solving this, we get: \[ x = 0 \]So, the vertical asymptote is at \(x = 0\).
02
- Find Horizontal Asymptotes
Horizontal asymptotes are determined by the degrees of the polynomial in the numerator and the denominator. The degree of the numerator is 1 and the degree of the denominator is 2. Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is y = 0.
03
- Find x-intercepts
The x-intercepts occur where the numerator is zero. For \(f(x)=\frac{x+1}{x^{2}}\), set the numerator equal to zero: \[ x + 1 = 0 \]Solving this, we get: \[ x = -1 \]So, the x-intercept is \( (-1, 0) \).
04
- Find y-intercepts
The y-intercepts occur where the function intersects the y-axis, which is when x = 0. However, for \(f(x)=\frac{x+1}{x^{2}}\), the function is undefined at x = 0 (vertical asymptote). Hence, there is no y-intercept.
05
- Sketch the Function
To sketch the function \(f(x)=\frac{x+1}{x^{2}}\), plot the x-intercept at (-1, 0). Add the vertical asymptote at x = 0, and the horizontal asymptote at y = 0. Observe the behavior of the function around these asymptotes. The function approaches these lines but never touches them.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertical Asymptotes
Vertical asymptotes are vertical lines where a rational function tends to infinity (positive or negative) as the input approaches a certain value. For the function \(f(x)=\frac{x+1}{x^{2}}\), we need to find where the denominator is zero. Instruction: Set the denominator equal to zero and solve for x. In this case, \(x^2 = 0\), so \(x = 0\). This tells us that there is a vertical asymptote at \(x = 0\). When sketching the graph, draw a dashed vertical line at \(x = 0\). It helps show where the function shoots to infinity.
Horizontal Asymptotes
Horizontal asymptotes indicate the value that a function approaches as the input becomes very large (positive or negative). To find these, compare the degrees of the numerator and the denominator of the rational function. In \(f(x) = \frac{x+1}{x^2}\), the degree of the numerator is 1 and the degree of the denominator is 2. Since the degree of the denominator is higher, the horizontal asymptote is at \(y = 0\). This means the function approaches the x-axis as \(x\) gets very large or very small. When sketching the graph, add a dashed horizontal line along \(y = 0\) to represent this.
x-intercepts
The x-intercepts of a function are the points where the graph crosses the x-axis. These occur where the numerator of the rational function is zero, provided the corresponding point does not make the denominator zero. For \(f(x) = \frac{x+1}{x^2}\), set the numerator equal to zero: \(x + 1 = 0\). Solving this, we find \(x = -1\). Therefore, there is an x-intercept at the point \((-1, 0)\). Include this point when sketching the graph.
y-intercepts
The y-intercepts of a function are the points where the graph crosses the y-axis. These occur at \(x = 0\). For \(f(x) = \frac{x+1}{x^2}\), substitute \(x = 0\) into the function. However, since the denominator becomes zero at \(x = 0\), the function is undefined here. Therefore, \(f(x)\) has no y-intercept. Remember this when sketching the graph and note that there’s no point where the graph crosses the y-axis.
Graph Sketching
To sketch the graph of the function \(f(x) = \frac{x+1}{x^2}\), follow these steps:
- Mark the vertical asymptote at \(x = 0\) with a dashed line.
- Mark the horizontal asymptote at \(y = 0\) with a dashed line.
- Plot the x-intercept at \((-1, 0)\).