Let's find the horizontal asymptotes of the function: \[ f(x) = \frac{4x^2 + \sqrt{4x^4 + x}}{2x^2 - 2} \] ### Step 1: Analyze the degrees - **Numerator:** The highest degree inside the numerator is from \(\sqrt{4x^4}\), which is \(2x^2\), so the numerator approaches \(\sim 2x^2\) for large \(|x|\). - **Denominator:** The highest degree term is \(2x^2\). ### Step 2: Dominant terms for large \(|x|\) Focus on the leading terms as \(x \to \pm\infty\): - \(4x^2 + \sqrt{4x^4 + x} \approx 4x^2 + 2x^2 = 6x^2\) - \(2x^2 - 2 \approx 2x^2\) Thus, as \(x \to \pm\infty\): \[ f(x) \approx \frac{6x^2}{2x^2} = 3 \] But let's be exact with the square root: \[ \sqrt{4x^4 + x} = |x^2| \sqrt{4 + \frac{1}{x^3}} \] As \(x \to \infty\), \(\sqrt{4 + \frac{1}{x^3}} \to 2\), so \(\sqrt{4x^4 + x} \to 2x^2\). So, as \(x \to \infty\): \[ f(x) \to \frac{4x^2 + 2x^2}{2x^2} = \frac{6x^2}{2x^2} = 3 \] But let's check as \(x \to -\infty\): For large negative \(x\), \(x^4\) is positive, but \(x\) is negative. The dominant term inside the root is still \(4x^4\), so: \[ \sqrt{4x^4 + x} \approx 2|x^2| = 2x^2 \text{ (since } x^2 \text{ is always positive)} \] So, same calculation applies, and as \(x \to -\infty\): \[ f(x) \to 3 \] ### Step 3: Double-check for negative side For very large negative \(x\): \[ \sqrt{4x^4 + x} \approx 2|x^2| = 2x^2 \] So answer is the same. ### Step 4: **Final Answer** The horizontal asymptote is: \[ y = 3 \] **But in the answer choices, "y = 3" is not listed. Let's check for possible errors or reconsider.** #### Look at the answer choices: - \(y = -1\) and \(y = 1\) - No horizontal asymptotes - \(y = 2\) only - \(y = 1\) only #### Let's try factoring out \(x^2\) from numerator and denominator for large \(x\): \[ f(x) = \frac{4x^2 + \sqrt{4x^4 + x}}{2x^2 - 2} = \frac{x^2 \left[4 + \sqrt{4 + \frac{1}{x^3}}\right]}{2x^2 (1 - \frac{1}{x^2})} \] So for large \(x\): \[ f(x) \to \frac{4 + 2}{2} = 3 \] **Again, 3 is not an option.** #### Let's check for possible sign errors: For \(x \to -\infty\), the square root is positive, so no sign change. #### But, let's try \(y = 1\): Let's analyze the behavior for very large and very small \(x\): - For very large positive/negative \(x\), \(f(x) \to 3\). This suggests an error in the choices or in the transcription of the problem. Based on the math, the **horizontal asymptote is \(y = 3\)**, **which is not given as an option**. **Please double-check the problem statement or the answer choices.** ### **Summary:** The correct horizontal asymptote for the given function is: \[ \boxed{y = 3} \] but this is not among the answer choices. If you must select from the given options, none are correct according to standard asymptote analysis.