Find the Slant Asymptote of \( f(x)=\left(4 x^{2}-16\right) /(x-3) \)

To find the **slant (oblique) asymptote** of the function \( f(x) = \frac{4x^2 - 16}{x - 3} \), follow these steps: ### Step 1: Check the Degrees A slant asymptote exists when the degree of the numerator is exactly one more than the degree of the denominator. - **Numerator:** \(4x^2 - 16\) has degree **2**. - **Denominator:** \(x - 3\) has degree **1**. Since \(2 = 1 + 1\), a slant asymptote exists. ### Step 2: Perform Polynomial Long Division Divide \(4x^2 - 16\) by \(x - 3\): 1. **Divide the leading terms:** \[ \frac{4x^2}{x} = 4x \] 2. **Multiply and subtract:** \[ (4x) \times (x - 3) = 4x^2 - 12x \] \[ (4x^2 - 16) - (4x^2 - 12x) = 12x - 16 \] 3. **Repeat the division:** \[ \frac{12x}{x} = 12 \] \[ 12 \times (x - 3) = 12x - 36 \] \[ (12x - 16) - (12x - 36) = 20 \] 4. **Combine the results:** \[ f(x) = 4x + 12 + \frac{20}{x - 3} \] ### Step 3: Identify the Slant Asymptote The slant asymptote is the linear part of the division result, ignoring the remainder term \(\frac{20}{x - 3}\). \[ \text{Slant Asymptote: } y = 4x + 12 \] **Final Answer:** The slant asymptote is the line y = 4 x + 12.