Answer is: option2
\( y = -\sqrt{3} \)Solution:
We are given the function:
\( f(x) = \frac{-3x^2}{\sqrt{3x^4 + 1}} \)
We are to find the horizontal asymptote of this function, i.e.,
\( \lim_{x \to \infty} f(x) \quad \text{and} \quad \lim_{x \to -\infty} f(x) \)
As \( x \to \infty \) or \( x \to -\infty \), the \( +1 \) in the denominator becomes negligible compared to \( 3x^4 \), so:
\[ f(x) \approx \frac{-3x^2}{\sqrt{3x^4}} = \frac{-3x^2}{x^2 \cdot \sqrt{3}} = \frac{-3}{\sqrt{3}} = -\sqrt{3} \]
So the horizontal asymptote is:
\( y = -\sqrt{3} \)
Final Answer: (B) \( y = -\sqrt{3} \)