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Magazine Chapter 4: Problem 34
Sketch the graph of each rational function. Specify the intercepts and the asymptotes. (a) \(f(x)=(x-1)(x+2.75) /[(x+1)(x+3)]\) (b) \(g(x)=(x-1)(x+3.25) /[(x+1)(x+3)]\) [Compare the graphs you obtain in parts (a) and (b). Notice how a relatively small change in one of the constants can radically alter the graph.]
Short Answer
Expert verified
The graphs of \(f(x)\) and \(g(x)\) have similar asymptotes but differ in x-intercepts, causing a shift in their plots.
Step by step solution
01
Identify the numerator and denominator
For both functions, let's identify the numerator and denominator. For \( f(x) \), the numerator is \((x-1)(x+2.75)\) and the denominator is \((x+1)(x+3)\). Similarly, for \( g(x) \), the numerator is \((x-1)(x+3.25)\) and the denominator is \((x+1)(x+3)\).
02
Determine the x-intercepts
The x-intercepts occur when the numerator is zero. For \( f(x) \), solve \((x-1)(x+2.75) = 0 \), giving x-intercepts at \(x=1\) and \(x=-2.75\). For \(g(x)\), solve \((x-1)(x+3.25) = 0\), giving x-intercepts at \(x=1\) and \(x=-3.25\).
03
Determine the y-intercepts
The y-intercept occurs when \(x=0\). Substitute \(x=0\) into \( f(x) \) and \( g(x) \):- \( f(0) = \frac{(0-1)(0+2.75)}{(0+1)(0+3)} = \frac{-2.75}{3} = -0.9167\)- \( g(0) = \frac{(0-1)(0+3.25)}{(0+1)(0+3)} = \frac{-3.25}{3} = -1.0833\)
04
Identify vertical asymptotes
Vertical asymptotes occur where the denominator is zero and the numerator is not. For both \( f(x) \) and \( g(x) \), set \((x+1)(x+3) = 0\), leading to vertical asymptotes at \(x=-1\) and \(x=-3\).
05
Identify horizontal asymptotes
Horizontal asymptotes for rational functions occur by analyzing the degrees of the polynomials in the numerator and denominator. Both functions have the same degree in the numerator and denominator (degree 2). Thus, the horizontal asymptote is \(y=\frac{1}{1}=1\).
06
Sketch the graphs
For \( f(x) \), plot x-intercepts at \( x=1 \) and \( x=-2.75 \), y-intercept at \( y=-0.9167 \), vertical asymptotes at \( x=-1 \) and \( x=-3 \), and horizontal asymptote at \( y=1 \).For \( g(x) \), plot x-intercepts at \( x=1 \) and \( x=-3.25 \), y-intercept at \( y=-1.0833 \), vertical asymptotes at \( x=-1 \) and \( x=-3 \), and horizontal asymptote at \( y=1 \).
07
Compare the graphs
Both functions have similar structures but differ in x-intercepts and y-intercepts, influenced by the variation in the constants. This slight change alters the location of the x and y-intercepts, shifting part of \(g(x)\) slightly to the left compared to \(f(x)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding x-intercepts in Rational Functions
In a rational function, the x-intercepts are the points where the graph crosses the x-axis. These occur where the numerator of the rational function equals zero, provided that the denominator is not zero at those points. This is because any fraction is zero when the numerator is zero and the denominator is not zero.
For instance, consider the function
For instance, consider the function
- \(f(x) = \frac{(x-1)(x+2.75)}{(x+1)(x+3)}\). Here, to find the x-intercepts, set the numerator \((x-1)(x+2.75) = 0\).
- This gives the solutions \(x = 1\) and \(x = -2.75\), as these values make the numerator zero.
- For \(g(x) = \frac{(x-1)(x+3.25)}{(x+1)(x+3)}\), solving \((x-1)(x+3.25) = 0\) reveals that the x-intercepts are at \(x = 1\) and \(x = -3.25\).
Finding y-intercepts of Rational Functions
The y-intercept of a function is the point where the graph intersects the y-axis. This occurs when the value of \(x\) is 0. For rational functions, the y-intercept can be found by substituting \(x = 0\) into the function and simplifying.
Consider the functions
Consider the functions
- \(f(x) = \frac{(x-1)(x+2.75)}{(x+1)(x+3)}\). The y-intercept is obtained by evaluating \(f(0)\), which results in \(\frac{(0-1)(0+2.75)}{(0+1)(0+3)} = \frac{-2.75}{3} = -0.9167\).
- For \(g(x) = \frac{(x-1)(x+3.25)}{(x+1)(x+3)}\) at \(x=0\), we find \(g(0) = \frac{(0-1)(0+3.25)}{(0+1)(0+3)} = \frac{-3.25}{3} = -1.0833\).
Exploring Vertical Asymptotes in Rational Functions
Vertical asymptotes in rational functions appear where the denominator equals zero, while the numerator remains non-zero. They are represented by vertical lines on the graph, often seen where the function tends to infinity or a sharp deviation.
For the functions at hand:
For the functions at hand:
- Both \(f(x)\) and \(g(x)\) have the same denominator. To find the vertical asymptotes, solve \((x+1)(x+3) = 0\).
- This results in \(x = -1\) and \(x = -3\) as points where the denominator is zero.
Analyzing Horizontal Asymptotes of Rational Functions
Horizontal asymptotes are lines parallel to the x-axis that the graph of a function may approach but never touch. They describe the function鈥檚 behavior as \(x\) goes to infinity or negative infinity. For rational functions, these can be determined by comparing the degrees of the polynomials in the numerator and denominator.
In both functions discussed here:
In both functions discussed here:
- The degree of the numerator is 2, as is the degree of the denominator.
- Because they are of equal degree, the horizontal asymptote is found by dividing the leading coefficients of the numerator and denominator.
- This results in a horizontal asymptote of \(y = \frac{1}{1} = 1\).
