Find the horizontal and vertical asymptotes of the curve. You may want to use a graphing calculator (or computer) to check your work by graphing the curve and estimating the asymptotes. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.)

*y* = 

*x*3 − *x*

*x*2 − 4*x* + 3
 
*x*=

1,3
 
 *y*=

x+4


We are given the rational function:

\[
y = \frac{x^3 - x}{x^2 - 4x + 3}
\]

We need to find the horizontal and vertical asymptotes of this curve. Let's go step by step.

### Step 1: Vertical Asymptotes

Vertical asymptotes occur where the denominator of the rational function is zero (and the numerator is not zero at the same point). To find the vertical asymptotes, we solve:

\[
x^2 - 4x + 3 = 0
\]

This is a quadratic equation, which we can factor:

\[
(x - 1)(x - 3) = 0
\]

So, the vertical asymptotes are at \(x = 1\) and \(x = 3\).

### Step 2: Horizontal Asymptotes

To find horizontal asymptotes, we look at the degrees of the numerator and denominator. The degree of the numerator is 3 (since the highest power of \(x\) is \(x^3\)), and the degree of the denominator is 2 (since the highest power of \(x\) is \(x^2\)).

Since the degree of the numerator is higher than the degree of the denominator, **there is no horizontal asymptote**. Instead, there is an **oblique (slant) asymptote**, which can be found by performing polynomial long division.

### Step 3: Oblique (Slant) Asymptote

To find the slant asymptote, we divide \(x^3 - x\) by \(x^2 - 4x + 3\) using long division. The result of this division will give us the equation of the oblique asymptote.

#### Polynomial Long Division:

\[
\frac{x^3 - x}{x^2 - 4x + 3}
\]

1. Divide \(x^3\) by \(x^2\), which gives \(x\).
2. Multiply \(x\) by \(x^2 - 4x + 3\), which gives \(x^3 - 4x^2 + 3x\).
3. Subtract this from \(x^3 - x\), which gives \(3x^2 - 4x\).
4. Divide \(3x^2\) by \(x^2\), which gives \(3\).
5. Multiply \(3\) by \(x^2 - 4x + 3\), which gives \(3x^2 - 12x + 9\).
6. Subtract this from \(3x^2 - 4x\), which gives \(8x - 9\).

The quotient is \(x + 3\), and the remainder is \(\frac{8x - 9}{x^2 - 4x + 3}\). As \(x\) becomes large, the remainder approaches 0, so the slant asymptote is:

\[
y = x + 3
\]

### Final Answers:
- **Vertical Asymptotes:** \(x = 1, 3\)
- **Horizontal Asymptotes:** DNE (but there is a slant asymptote)
- **Slant Asymptote:** \(y = x + 3\)

Would you like a detailed explanation of any step?

### 5 Related Questions:
1. How do you determine when a rational function has a horizontal asymptote?
2. What is the difference between vertical and horizontal asymptotes?
3. How do you perform polynomial long division for rational functions?
4. How do slant asymptotes differ from horizontal asymptotes?
5. Can a rational function have both horizontal and slant asymptotes?

### Tip:
When the degree of the numerator exceeds the degree of the denominator by exactly one, expect a slant asymptote instead of a horizontal one.

Find the horizontal and vertical asymptotes of the curve. You may want to use a graphing calculator (or computer) to check your work by graphing the curve and estimating the asymptotes. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.) y = (x^3 - x)/(x^2 - 4x + 3)

Learn how to find the horizontal and vertical asymptotes of the rational function y = (x^3 - x)/(x^2 - 4x + 3). This solution explains key concepts such as vertical asymptotes, horizontal asymptotes, and slant asymptotes using polynomial long division. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation