(2) Find vertical and horizontal asymptotes of the following functions. Show
your work. Evaluate all the necessary limits.
(i)
(ii)

Ask by Chandler Turnbull. in Canada

Jan 26,2025

Question

(2) Find vertical and horizontal asymptotes of the following functions. Show
your work. Evaluate all the necessary limits.
(i)
(ii)

Ask by Chandler Turnbull. in Canada

Jan 26,2025

UpStudy AI · Accepted Answer

Let's analyze each function to find their vertical and horizontal asymptotes by evaluating the necessary limits and performing algebraic simplifications where needed.

---

### **Problem (i)**

**Function:**
$$ f(x) = \frac{2x - 1}{3x^2 + 5x - 2} $$

#### **1. Vertical Asymptotes**

Vertical asymptotes occur where the denominator is zero (since the function is undefined there) and the numerator is not zero at those points.

1. **Find the roots of the denominator:**
   $$
   3x^2 + 5x - 2 = 0
   $$
   
2. **Solve the quadratic equation using the quadratic formula:**
   $$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$
   Here, $$ a = 3 $$, $$ b = 5 $$, and $$ c = -2 $$.
   
3. **Calculate the discriminant:**
   $$
   \Delta = b^2 - 4ac = 25 + 24 = 49
   $$
   
4. **Find the roots:**
   $$
   x = \frac{-5 \pm \sqrt{49}}{6} = \frac{-5 \pm 7}{6}
   $$
   - $$ x = \frac{2}{6} = \frac{1}{3} $$
   - $$ x = \frac{-12}{6} = -2 $$
   
5. **Check the numerator at these points:**
   $$
   \text{At } x = \frac{1}{3}: \quad 2\left(\frac{1}{3}\right) - 1 = \frac{2}{3} - 1 = -\frac{1}{3} \neq 0
   $$
   $$
   \text{At } x = -2: \quad 2(-2) - 1 = -4 - 1 = -5 \neq 0
   $$
   
   Since the numerator is not zero at $$ x = \frac{1}{3} $$ and $$ x = -2 $$, these are indeed vertical asymptotes.

**Vertical Asymptotes:**
$$
x = \frac{1}{3} \quad \text{and} \quad x = -2
$$

#### **2. Horizontal Asymptote**

To determine the horizontal asymptote, compare the degrees of the numerator and the denominator.

- **Degree of the numerator (2x - 1):** 1
- **Degree of the denominator (3x² + 5x - 2):** 2

Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is:
$$
y = 0
$$

---

### **Problem (ii)**

**Function:**
$$ f(x) = \frac{-x^3 + 2x^2}{x^4 - 4x^3 + 4x^2} $$

#### **1. Simplify the Function**

First, factor both the numerator and the denominator:

1. **Factor the numerator:**
   $$
   -x^3 + 2x^2 = -x^2(x - 2)
   $$
   
2. **Factor the denominator:**
   $$
   x^4 - 4x^3 + 4x^2 = x^2(x^2 - 4x + 4) = x^2(x - 2)^2
   $$
   
3. **Simplify the function by canceling common factors:**
   $$
   f(x) = \frac{-x^2(x - 2)}{x^2(x - 2)^2} = \frac{-1}{x - 2}
   $$
   
   **Note:** The factors $$ x^2 $$ and $$ (x - 2) $$ in the numerator and denominator cancel out, except for one $$ (x - 2) $$ in the denominator. However, $$ x = 0 $$ causes both numerator and denominator to be zero, indicating a **removable discontinuity** (a hole) at $$ x = 0 $$, not a vertical asymptote.

#### **2. Vertical Asymptote**

After simplification, the function becomes:
$$
f(x) = \frac{-1}{x - 2}
$$

The denominator is zero when:
$$
x - 2 = 0 \quad \Rightarrow \quad x = 2
$$

**Behavior near $$ x = 2 $$:**
$$
\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{-1}{x - 2} = -\infty
$$
$$
\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{-1}{x - 2} = +\infty
$$

Since the limits approach infinity from different directions, there is a vertical asymptote at $$ x = 2 $$.

**Vertical Asymptote:**
$$
x = 2
$$

#### **3. Horizontal Asymptote**

Compare the degrees of the original numerator and denominator:

- **Degree of the numerator (-x³ + 2x²):** 3
- **Degree of the denominator (x⁴ - 4x³ + 4x²):** 4

Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is:
$$
y = 0
$$

---

### **Summary of Asymptotes**

1. **Function (i):**
   - **Vertical Asymptotes:** $$ x = \frac{1}{3} $$ and $$ x = -2 $$
   - **Horizontal Asymptote:** $$ y = 0 $$

2. **Function (ii):**
   - **Vertical Asymptote:** $$ x = 2 $$
   - **Horizontal Asymptote:** $$ y = 0 $$
**Asymptotes:** 1. **Function (i):** - **Vertical Asymptotes:** $$ x = \frac{1}{3} $$ and $$ x = -2 $$ - **Horizontal Asymptote:** $$ y = 0 $$ 2. **Function (ii):** - **Vertical Asymptote:** $$ x = 2 $$ - **Horizontal Asymptote:** $$ y = 0 $$

(2) Find vertical and horizontal asymptotes of the following functions. Show
your work. Evaluate all the necessary limits.
(i)
(ii)

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(2) Find vertical and horizontal asymptotes of the following functions. Show your work. Evaluate all the necessary limits. (i) (ii)

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