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Give an example of a rational function that has vertical asymptote \(x=3 .\) Now give an example of one that has vertical asymptote \(x=3\) and horizontal asymptote \(y=2 .\) Now give an example of a rational function with vertical asymptotes \(x=1\) and \(x=-1,\) horizontal asymptote \(y=0,\) and \(x\) -intercept 4.

Short Answer

Expert verified
Examples: \(\frac{1}{x-3}\), \(\frac{2x}{x-3}\), and \(\frac{x-4}{(x-1)(x+1)}\).

Step by step solution

01

Identifying Vertical Asymptotes

The vertical asymptote(s) of a rational function occur where the denominator is zero but the numerator is not zero. For a rational function with a vertical asymptote at \(x=3\), we need a factor of \((x-3)\) in the denominator. An example function is \( f(x) = \frac{1}{x-3} \).
02

Adding a Horizontal Asymptote

Horizontal asymptotes depend on the degrees of the polynomials in the numerator and denominator. To have a horizontal asymptote at \(y=2\) when there is also a vertical asymptote at \(x=3\), the degrees of the numerator and the denominator should be equal. A possible function is \( f(x) = \frac{2x}{x-3}\).
03

Building a Function with Multiple Vertical Asymptotes

For vertical asymptotes at \(x=1\) and \(x=-1\), the denominator should have factors \((x-1)(x+1)\). A suitable function is \( f(x) = \frac{x}{(x-1)(x+1)} \) which also has a horizontal asymptote \(y=0\) because the degree of the numerator is less than the degree of the denominator.
04

Adjusting for an x-intercept

To have an \(x\)-intercept at 4, the numerator of the rational function should be zero when \(x=4\). Therefore, the numerator should have a factor \((x-4)\). Combining this with the requirement for vertical asymptotes at \(x=1\) and \(x=-1\) and horizontal asymptote \(y=0\), we have the function \( f(x) = \frac{x-4}{(x-1)(x+1)} \).

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Key Concepts

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

Vertical Asymptotes
In rational functions, vertical asymptotes are lines where the function tends to infinity. These occur when the denominator of the function equals zero, but the numerator does not.
Imagine a rational function like a fraction, where the denominator determines the vertical asymptotes. If you have a function such as \[ f(x) = \frac{1}{x-3} \], it gains a vertical asymptote at \( x = 3 \) because when you substitute \( x = 3 \) into the denominator, it results in zero, thus making the function undefined.
For multiple vertical asymptotes, factors of those specific \( x \)-values must exist in the denominator. For example, using \[ f(x) = \frac{x}{(x-1)(x+1)} \], you create vertical asymptotes at \( x = 1 \) and \( x = -1 \).
Horizontal Asymptotes
Horizontal asymptotes provide us with valuable information on the end behavior of rational functions. They indicate the value that \( y \) approaches as \( x \) moves towards infinity. The degree or the highest power of \( x \) in the numerator and the denominator determines their presence.
When the degrees of the numerator and the denominator match, the horizontal asymptote is the ratio of their leading coefficients. For instance, in \[ f(x) = \frac{2x}{x-3} \], the degrees are the same (both 1), giving us a horizontal asymptote at \( y = 2 \).
Conversely, if the numerator's degree is less than that of the denominator, like in \[ f(x) = \frac{x}{(x-1)(x+1)} \], then \( y = 0 \)becomes the asymptote, as occurs when the denominator manages growth faster than the numerator.
X-Intercepts
A rational function achieves an \( x \)-intercept when its value is zero, which occurs when the numerator is equal to zero. This is because any number divided by zero results in zero, but the converse isn't true since dividing by zero is undefined.
To place an \( x \)-intercept at a particular point, we include that \( x \)-value as a factor in the numerator. For a function needing an intercept at \( x = 4 \), ensuring \( (x-4) \) is present in the numerator, such as in \[ f(x) = \frac{x-4}{(x-1)(x+1)} \], will meet this requirement. This function will have \( x \)-intercept \( (4,0) \), while simultaneously preserving vertical asymptotes at \( x = 1 \) and \( x = -1 \).
Denominator of a Function
The denominator plays a crucial part in shaping the behavior and characteristics of rational functions, like where vertical asymptotes and domain restrictions occur. \( x \)-values causing the denominator to be zero lead to undefined segments in the function – these often manifest as vertical asymptotes.
Consider that manipulating the denominator can determine the number and location of vertical asymptotes. For example, \[ (x-3) \]results in an asymptote at \( x=3 \), and treating it as \[ (x-1)(x+1) \]generates vertical asymptotes at \( x=1 \) and \( x=-1 \).
The denominator also influences the horizontal asymptotes and end behavior as its degree, in comparison to the numerator, dictates the asymptote's position in the \( y \)-axis, making it integral to understanding rational functions' overall structure.

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Most popular questions from this chapter

The real solutions of the given equation are rational. List all possible rational roots using the Rational Zeros Theorem, and then graph the polynomial in the given viewing rectangle to determine which values are actually solutions. (All solutions can be seen in the given viewing rectangle.) $$x^{4}-5 x^{2}+4=0 ; \quad[-4,4] \text { by }[-30,30]$$

Graph the rational function \(f\) and determine all vertical asymptotes from your graph. Then graph \(f\) and \(g\) in a sufficiently large viewing rectangle to show that they have the same end behavior. $$f(x)=\frac{-x^{4}+2 x^{3}-2 x}{(x-1)^{2}}, g(x)=1-x^{2}$$

The real solutions of the given equation are rational. List all possible rational roots using the Rational Zeros Theorem, and then graph the polynomial in the given viewing rectangle to determine which values are actually solutions. (All solutions can be seen in the given viewing rectangle.) $$3 x^{3}+8 x^{2}+5 x+2=0 ; \quad[-3,3] \text { by }[-10,10]$$

Find all rational zeros of the polynomial, and then find the irrational zeros, if any. Whenever appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes' Rule of Signs, the quadratic formula, or other factoring techniques. $$P(x)=2 x^{4}+15 x^{3}+31 x^{2}+20 x+4$$

Graph the rational function and find all vertical asymptotes, \(x\)- and \(y\)-intercepts, and local extrema, correct to the nearest decimal. Then use long division to find a polynomial that has the same end behavior as the rational function, and graph both functions in a sufficiently large viewing rectangle to verify that the end behaviors of the polynomial and the rational function are the same. $$y=\frac{x^{4}}{x^{2}-2}$$

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