To find the vertical asymptotes of the function

$y = \frac{2e^{2x} + e^{-2x}}{2e^{2x} - 3e^{-2x}},$

we need to identify the values of $x$ that make the denominator zero, since vertical asymptotes occur where the function approaches infinity.

Step 1: Set the Denominator to Zero

Set the denominator equal to zero:

$2e^{2x} - 3e^{-2x} = 0.$

Step 2: Clear the Exponential

To solve for $x$, multiply both sides by $e^{2x}$ (which is always positive):

$2e^{4x} - 3 = 0.$

Step 3: Solve for $e^{4x}$

Rearranging gives:

$2e^{4x} = 3,$

which simplifies to:

$e^{4x} = \frac{3}{2}.$

Step 4: Take the Natural Logarithm

Taking the natural logarithm of both sides:

$4x = \ln\left(\frac{3}{2}\right).$

Step 5: Solve for $x$

Finally, divide by 4:

$x = \frac{1}{4} \ln\left(\frac{3}{2}\right).$

Conclusion

The vertical asymptote of the function occurs at:

$x = \frac{1}{4} \ln\left(\frac{3}{2}\right).$

If you need further details or have any questions, let me know! Here are some related questions to consider:

1. What does the vertical asymptote indicate about the behavior of the function near that point?
2. How can you determine horizontal asymptotes for this function?
3. What happens to the function values as $x$ approaches the vertical asymptote from the left and right?
4. Can you find the intercepts of this function?
5. How would you graph this function, including the asymptotes?

Tip: When analyzing rational functions, always check both the numerator and the denominator for potential zeros and points of discontinuity.