To find the horizontal asymptote of the function $g(x) = \frac{2x^2 - 3^x}{4^x - 5x^2}$, we need to analyze the behavior of the function as $x$ approaches infinity ($x \to \infty$) or negative infinity ($x \to -\infty$).

Step 1: Analyze the growth rates of the terms

• $2x^2$ grows quadratically.
• $3^x$ grows exponentially.
• $4^x$ grows exponentially.
• $5x^2$ grows quadratically.

Since exponential functions grow much faster than polynomial functions as $x$ becomes large, the terms involving $3^x$ and $4^x$ will dominate as $x \to \infty$.

Step 2: Simplify the function for large $x$

For large $x$, we can approximate $g(x)$ by considering only the leading exponential terms: $g(x) \approx \frac{-3^x}{4^x} = \frac{-\left(3/4\right)^x}{1}$ As $x \to \infty$, the term $\left(\frac{3}{4}\right)^x$ approaches 0 because $\frac{3}{4} < 1$. Therefore: $g(x) \to 0 \text{ as } x \to \infty$

Step 3: Conclusion

The horizontal asymptote of the function $g(x)$ is: $y = 0$

Would you like more details or have any questions?

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Here are 5 related questions:

1. What is the vertical asymptote of $g(x)$?
2. How do the horizontal asymptotes change if the exponents are altered?
3. Can we find the horizontal asymptote if $g(x)$ had logarithmic functions?
4. What happens to the asymptotes if the coefficients of $x^2$ terms change?
5. How do we determine slant (oblique) asymptotes?

Tip: When finding horizontal asymptotes, focus on the terms that grow the fastest as $x$ approaches infinity or negative infinity.