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Graph the rational functions in Exercises \(27-38 .\) Include the graphs and equations of the asymptotes and dominant terms. $$ y=\frac{x^{2}}{x-1} $$

Short Answer

Expert verified
Vertical asymptote at \( x = 1 \); oblique asymptote: \( y = x + 1 \); x-intercept at \( (0, 0) \).

Step by step solution

01

Identify the Function Components

The given function is a rational function \( y = \frac{x^2}{x-1} \). It consists of a polynomial numerator \( x^2 \) and a polynomial denominator \( x - 1 \). To analyze this function, we need to find the asymptotes, intercepts, and identify any dominant terms.
02

Find Vertical Asymptotes

Vertical asymptotes occur when the denominator of the function is equal to zero. Set \( x - 1 = 0 \) and solve for \( x \), which gives \( x = 1 \). Therefore, there is a vertical asymptote at \( x = 1 \).
03

Find Horizontal Asymptotes

To find horizontal asymptotes, compare the degrees of the numerator and the denominator. The degree of the numerator (\( x^2 \)) is 2, and the degree of the denominator (\( x - 1 \)) is 1. Since the degree of the numerator is greater, there is no horizontal asymptote. There will be an oblique asymptote instead.
04

Identify Oblique Asymptote

The oblique asymptote occurs because the degree of the numerator is exactly one more than the degree of the denominator. Perform polynomial long division on \( x^2 \div (x-1) \), which results in \( x + 1 \) with a remainder of \( 1 \). Hence, the equation of the oblique asymptote is \( y = x + 1 \).
05

Find the X-Intercepts

To find the x-intercepts, set the numerator equal to zero and solve for \( x \). For \( x^2 = 0 \), \( x = 0 \). Thus, there is an intercept at \( (0, 0) \).
06

Find the Y-Intercept

The y-intercept occurs where the graph crosses the y-axis, which is when \( x = 0 \). Substituting \( x = 0 \) into the function \( y = \frac{0^2}{0-1} = 0 \), confirms that the y-intercept is also at \( (0, 0) \).
07

Sketch the Graph

To sketch the graph, plot the x-intercept \((0,0)\), and the vertical asymptote at \( x=1 \). The oblique asymptote \( y=x+1 \) acts as a guide for the end-behavior of the graph. The graph approaches the vertical asymptote as it nears \( x = 1 \) and aligns closely with the oblique asymptote for large positive and negative values of \( x \).

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Key Concepts

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

Vertical Asymptotes
When graphing rational functions, vertical asymptotes play a crucial role. They occur at points where the function is undefined, leading to a dramatic change in direction on the graph. To identify vertical asymptotes, you need to set the denominator equal to zero. For example, in the function \( y = \frac{x^2}{x-1} \), setting the denominator \( x-1 \) equal to zero gives us \( x = 1 \).
This means the graph will approach but never touch the vertical line \( x = 1 \). As \( x \) gets close to 1 from the left or right, the function's values will increase or decrease without bound (towards infinity).
This is why it seems like the graph is splitting at this point, creating a line of discontinuity known as a vertical asymptote. Always look for points where the denominator reaches zero to find these asymptotes.
Horizontal Asymptotes
Horizontal asymptotes differ from vertical ones because they concern the end behavior of a graph, not any particular x-value where the function is undefined. They exist in rational functions when the degree of the numerator is less than or equal to the degree of the denominator.
To find a horizontal asymptote, you compare these degrees. In our example, \( y = \frac{x^2}{x-1} \), the numerator has a degree of 2, while the denominator has a degree of 1. Since the numerator's degree is greater, there is no horizontal asymptote. Instead, you look for an oblique asymptote. If the degrees were equal, you would simplify the leading coefficients to find the horizontal asymptote, as the value the function approaches as \( x \) approaches infinity.
Oblique Asymptotes
Oblique asymptotes, sometimes called slant asymptotes, appear when the degree of the numerator is exactly one more than the degree of the denominator in a rational function. Instead of a flat line like a horizontal asymptote, these are diagonal.
To find an oblique asymptote, you perform polynomial long division. For \( y = \frac{x^2}{x-1} \), dividing gives \( y = x + 1 \) with some remainder that becomes insignificant for large values of \( x \). The function will approach the line \( y = x + 1 \) as \( x \) becomes very large in both positive and negative directions.
The oblique asymptote predicts how the function behaves at extreme values of \( x \), acting as a guide to where the tails of the graph will head, providing essential information on the overarching shape of the graph.

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Most popular questions from this chapter

Graphing Secant and Tangent Lines Use a CAS to perform the following steps for the functions in Exercises \(45-48 .\) a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\) b. Holding \(x_{0}\) fixed, the difference quotient $$ q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h} $$ at \(x_{0}\) becomes a function of the step size \(h .\) Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\) d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h=3,2\) and \(1 .\) Graph them together with \(f\) and the tangent line over the interval in part (a). $$ f(x)=x^{3}+2 x, \quad x_{0}=0 $$

In Exercises \(43-46\) , find a function that satisfies the given conditions and sketch its graph. (The answers here are are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.) $$ \lim _{x \rightarrow \pm \infty} k(x)=1, \lim _{x \rightarrow 1^{-}} k(x)=\infty, \text { and } \lim _{x \rightarrow 1^{+}} k(x)=-\infty $$

Each of the functions in Exercises \(57-60\) is given as the sum or difference of two terms. First graph the terms (with the same set of axes). Then, using these graphs as guides, sketch in the graph of the function. $$ y=\sec x+\frac{1}{x}, \quad-\frac{\pi}{2} < x < \frac{\pi}{2} $$

In Exercises \(11-18,\) find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$ f(x)=x^{2}+1, \quad(2,5) $$

In Exercises \(5-10\) , find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. $$ y=4-x^{2}, \quad(-1,3) $$

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