Written Activity 4

I. Identify the domain, range, x and y intercepts, vertical and horizontal asymptote (if no HA, Oblique Asymptote).



3. f(x) = (x ^ 2)/(x - 1)

Let's analyze the function \( f(x) = \frac{x^2}{x - 1} \) in detail:

### 1. **Domain**
The domain refers to the set of all possible input values (x-values) for the function. The denominator cannot be zero because division by zero is undefined.

\[
x - 1 \neq 0 \quad \Rightarrow \quad x \neq 1
\]

Thus, the domain is:
\[
\text{Domain}: (-\infty, 1) \cup (1, \infty)
\]

### 2. **Range**
To determine the range, we analyze the behavior of the function as \( x \to \infty \) and \( x \to 1 \). Since there is a vertical asymptote at \( x = 1 \), the function has different behaviors approaching 1 from the left and right. Additionally, since the numerator is quadratic and the denominator is linear, the function can take any real value as \( x \to \pm \infty \). Therefore, the function's range is:

\[
\text{Range}: (-\infty, \infty)
\]

### 3. **x-Intercept**
The x-intercepts occur when \( f(x) = 0 \), which happens when the numerator equals zero. In this case:

\[
x^2 = 0 \quad \Rightarrow \quad x = 0
\]

Thus, the x-intercept is:
\[
x\text{-intercept}: (0, 0)
\]

### 4. **y-Intercept**
The y-intercept occurs when \( x = 0 \). Plugging \( x = 0 \) into the function:

\[
f(0) = \frac{0^2}{0 - 1} = 0
\]

Thus, the y-intercept is:
\[
y\text{-intercept}: (0, 0)
\]

### 5. **Vertical Asymptote**
Vertical asymptotes occur where the function tends towards infinity, which happens when the denominator is zero. In this case:

\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

Thus, the vertical asymptote is:
\[
\text{Vertical Asymptote}: x = 1
\]

### 6. **Oblique (Slant) Asymptote**
Since the degree of the numerator (2) is one greater than the degree of the denominator (1), there is an oblique asymptote. To find the oblique asymptote, we perform polynomial division:

\[
\frac{x^2}{x - 1} = x + 1 + \frac{1}{x - 1}
\]

As \( x \to \infty \), the remainder term \( \frac{1}{x - 1} \) approaches zero, so the oblique asymptote is:

\[
\text{Oblique Asymptote}: y = x + 1
\]

### Summary:
- **Domain**: \( (-\infty, 1) \cup (1, \infty) \)
- **Range**: \( (-\infty, \infty) \)
- **x-intercept**: \( (0, 0) \)
- **y-intercept**: \( (0, 0) \)
- **Vertical Asymptote**: \( x = 1 \)
- **Oblique Asymptote**: \( y = x + 1 \)

Would you like more details on any step, or do you have further questions?

### 5 Related Questions:
1. What is the relationship between the degree of the numerator and denominator in determining asymptotes?
2. How can you determine if a function has a slant asymptote or a horizontal asymptote?
3. Why can't the function be defined at \( x = 1 \)?
4. How do oblique asymptotes affect the graph of a function?
5. What other methods can be used to find intercepts?

### Tip:
When dealing with rational functions, always check for vertical asymptotes by identifying when the denominator equals zero and for oblique asymptotes if the degree of the numerator exceeds that of the denominator by exactly one.

Identify the domain, range, x and y intercepts, vertical and horizontal asymptote (if no HA, Oblique Asymptote) for f(x) = (x^2)/(x - 1).

Explore the rational function f(x) = (x^2)/(x - 1) with a detailed analysis of its domain, range, x and y intercepts, vertical asymptote at x = 1, and oblique asymptote y = x + 1. This problem focuses on key algebraic concepts such as intercepts, asymptotes, and polynomial division, suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation