To find the vertical asymptotes of the function $y = \frac{x^2 + 1}{7x - 4x^2}$, we need to determine the values of $x$ that make the denominator equal to zero, since vertical asymptotes occur when the function approaches infinity (i.e., where the denominator is zero and the numerator is non-zero).

The denominator of the function is: $7x - 4x^2 = 0$

Let's solve this equation for $x$.

$7x = 4x^2$ $x(7 - 4x) = 0$

Setting each factor to zero:

1. $x = 0$
2. $7 - 4x = 0 \Rightarrow x = \frac{7}{4}$

Thus, the vertical asymptotes occur at: $x = 0, \quad x = \frac{7}{4}$

Final Answer:

The vertical asymptotes are at $x = 0$ and $x = \frac{7}{4}$.

Would you like any further explanation on these steps, or additional details?

Here are 5 related questions you might find helpful:

1. What are horizontal asymptotes, and how can they be found for rational functions?
2. How do we handle situations where the numerator and denominator both approach zero at the same value of $x$?
3. How does the degree of the numerator and denominator affect the asymptotes?
4. Can a rational function have holes in the graph, and how can they be identified?
5. How do slant asymptotes differ from vertical and horizontal asymptotes?

Tip: When solving for vertical asymptotes, always factor the denominator and check for common factors with the numerator—this can reveal points where the graph has holes instead of asymptotes!