(2) Find vertical and horizontal asymptotes of the following functions. Show your work. Evaluate all the necessary limits. (i) \( f(x)=\frac{2 x-1}{3 x^{2}+5 x-2} \) (ii) \( f(x)=\frac{-x^{3}+2 x^{2}}{x^{4}-4 x^{3}+4 x^{2}} \)

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:** \( y = 0 \)