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Write a rational function \(f\) that has the specified characteristics. (There are many correct answers.) (a) Vertical asymptote: \(x=2\) Horizontal asymptote: \(y=0\) Zero: \(x=1\) (b) Vertical asymptote: \(x=-1\) Horizontal asymptote: \(y=0\) Zero: \(x=2\) (c) Vertical asymptotes: \(x=-2, x=1\) Horizontal asymptote: \(y=2\) Zeros: \(x=3, x=-3\) (d) Vertical asymptotes: \(x=-1, x=2\) Horizontal asymptote: \(y=-2\) Zeros: \(x=-2, x=3\) (c) Vertical asymptotes: \(x=0, x=\pm 3\) Horizontal asymptote: \(y=3\) Zeros: \(x=-1, x=1, x=2\)

Short Answer

Expert verified
The respective rational functions are (a) \( f(x)=\frac{ x-1 }{ x-2 }\), (b) \(f(x)=\frac{ x-2 } {x+1 }\), (c) \(f(x)=2\frac{ (x-3)(x+3) } { (x+2)(x-1) }\), (d) \(f(x)=-2\frac{ (x+2)(x-3) } { (x+1)(x-2) }\) and (e) \(f(x)=3\frac{ (x+1)(x-1)(x-2) } { x(x^2-9) }\)

Step by step solution

01

Function for (a)

For the given characteristics, a suitable function could be \( f(x)=\frac{ x-1 }{ x-2 }\). Here, x=2 is a vertical asymptote, y=0 is the horizontal asymptote, and x=1 is a zero of the function.
02

Function for (b)

A function that meets the given criteria is \(f(x)=\frac{ x-2 } {x+1 }\). The vertical asymptote is x=-1, the horizontal asymptote is y=0, and x=2 is a zero of the function.
03

Function for (c)

The characteristics suggest the following function: \(f(x)=2\frac{ (x-3)(x+3) } { (x+2)(x-1) }\). Here, the function has vertical asymptotes at x=-2 and x=1, a horizontal asymptote at y=2, and zeros at x=3 and x=-3.
04

Function for (d)

The function that fits the specified features might be \(f(x)=-2\frac{ (x+2)(x-3) } { (x+1)(x-2) }\). The vertical asymptotes are x=-1 and x=2, the horizontal asymptote is y=-2, and the zeros are x=-2 and x=3.
05

Function for (e)

Finally, for these conditions we can propose the function \(f(x)=3\frac{ (x+1)(x-1)(x-2) } { x(x^2-9) }\). This function has vertical asymptotes at x=0 and x=\pm 3, a horizontal asymptote at y=3, and zeros at x=-1, x=1, and x=2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertical Asymptotes
Vertical asymptotes in rational functions denote the values of the variable where the function's value becomes infinitely large or small, giving a sense of where the graph of the function approaches but never touches.

For instance, in the given exercise where we have the function \( f(x)=\frac{x-1}{x-2} \), \( x=2 \) is the vertical asymptote. At this point, the denominator of the function equals zero, making the function undefined and the graph of the function tends toward infinity (or negative infinity) as \( x \) approaches the vertical asymptote from either side of the graph.

Remember, a function may have multiple vertical asymptotes, as seen in some of the given exercises where the denominators contain more than one factor that could be set to zero. Vertical asymptotes are crucial for sketching the graph of rational functions.
Horizontal Asymptotes
Horizontal asymptotes, on the other hand, refer to the line that the graph of a rational function approaches as the variable either goes to positive or negative infinity. It's a way to determine the end behavior of a function.

In the exercise solutions, the horizontal asymptotes are all represented by the equation \( y= \) some constant. This asymptote tells us the value that \( f(x) \) gets closer to, but doesn't actually reach, as \( x \) increases or decreases without bound. For instance, the function \( f(x)=\frac{x-2}{x+1} \) has a horizontal asymptote at \( y=0 \), indicating that as \( x \) becomes very large or very negative, the value of \( f(x) \) approaches zero.

The horizontal asymptotes depend largely on the degrees of the polynomials in the numerator and denominator of the function. If the degree of the numerator is equal to or less than the degree of the denominator, the horizontal asymptote will be at \( y=0 \); if the degrees are equal, the horizontal asymptote will be at \( y= \) the ratio of the leading coefficients.
Zeros of a Function
The zeros of a function are the values of the variable that make the function equal to zero. These are the points where the graph of the function will cross the horizontal axis.

For a rational function, zeros are found by setting the numerator equal to zero and solving for the variable. For example, in our function \( f(x)=\frac{x-2}{x+1} \), the numerator \( x-2 = 0 \) gives us the zero at \( x=2 \). Each zero translates into an \( x \)-intercept on the graph of the function.

Zeros are important for understanding the behavior and characteristics of the function, as well as for graphing. They provide key points through which the graph will pass, offering a framework for the rest of the graph. Typically, the behavior of rational functions near their zeros will depend on the order of the zero — if the zero has an even multiplicity, the graph will touch the axis and turn back; if the zero is of odd multiplicity, the graph will cross the axis.
Graphing Rational Functions
Graphing rational functions involves a step-by-step process that takes into account the zeros, vertical and horizontal asymptotes, and often other features such as y-intercepts or symmetry.

The rational functions provided in the exercise illustrate how knowing the zeros and asymptotes can form an initial 'skeleton' of the graph. To accurately sketch these functions, you start by plotting the asymptotes as dashed lines that the graph will never touch. Next, plot the zeros where the function intersects the \( x \)-axis.

After placing these foundational elements on the graph, it is possible to sketch the general shape of the function. It's important to note how the function behaves near the asymptotes and zeros: for example, whether it approaches the asymptotes from above or below, and how it passes through the zeros. By understanding these elements, graphing any rational function can become a manageable, step-by-step process.

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Most popular questions from this chapter

A rectangular package sent by a delivery service can have a maximum combined length and girth (perimeter of a cross section) of 120 inches (see figure). (a) Show that the volume of the package is given by the function \(V(x)=4 x^{2}(30-x)\) (b) Use a graphing utility to graph the function and approximate the dimensions of the package that yield a maximum volume. (c) Find values of \(x\) such that \(V=13,500 .\) Which of these values is a physical impossibility in the construction of the package? Explain.

Match the cubic function with the correct number of rational and irrational zeros. (a) Rational zeros: \(0 ; \quad\) Irrational zeros: 1 (b) Rational zeros: \(3 ;\) Irrational zeros: \(\mathbf{0}\) (c) Rational zeros: \(1 ; \quad\) Irrational zeros: 2 (d) Rational zeros: \(1 ;\) Irrational zeros: \(\mathbf{0}\) $$f(x)=x^{3}-2 x$$

Determine whether the statement is true or false. Justify your answer. A rational function can have infinitely many vertical asymptotes.

Write a rational function that has the specificd characteristics. (There are many correct answers.) (a) Vertical asymptote: \(x=-2\) Slant asymptote: \(y=x+1\) Zero of the function: \(x=2\) (b) Vertical asymptote: \(x=-4\) Slant asymptote: \(y=x-2\) Zero of the function: \(x=3\)

Use a graphing utility to compare the graphs of \(y_{1}\) and \(y_{2}.\) $$y_{1}=\frac{3 x^{3}-5 x^{2}+4 x-5}{2 x^{2}-6 x+7}, \quad y_{2}=\frac{3 x^{3}}{2 x^{2}}$$ Start with a viewing window of \(-5 \leq x \leq 5\) and \(-10 \leq y \leq 10,\) and then zoom out. Make a conjecture about how the graph of a rational function \(f\) is related to the graph of \(y=a_{n} x^{n} / b_{m} x^{m},\) where \(a_{n} x^{n}\) is the leading term of the numerator of \(f\) and \(b_{m} x^{m}\) is the leading term of the denominator of \(f.\)

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