To find the intersection points of the given function $y = \frac{x^3 - 1}{4x^2}$ with the $OX$-axis, we need to solve for where $y = 0$.

Steps:

1. Set $y = 0$: $\frac{x^3 - 1}{4x^2} = 0$

2. The numerator must equal zero for the fraction to be zero (since the denominator cannot be zero): $x^3 - 1 = 0$

3. Solve for $x$: $x^3 = 1 \quad \implies \quad x = 1$

4. At $x = 1$, substitute back into the function to confirm $y = 0$: $y = \frac{(1)^3 - 1}{4(1)^2} = \frac{0}{4} = 0$

Thus, the intersection point is $x = 1$ and $y = 0$.

Final Answer:

10 (written as $x = 1, y = 0$).

Do you want further explanation on any step?

Here are related questions to expand understanding:

1. How do we find the intersection points with the $OY$-axis instead?
2. What if the denominator also allowed a solution $x = 0$? (Asymptote discussion)
3. How does the behavior of $x^3$ and $x^2$ affect the graph near $x = 0$?
4. Could this function have vertical asymptotes? If so, where?
5. How would the graph behave at larger positive or negative values of $x$?

Tip: Always ensure the denominator in a rational function is non-zero to avoid undefined points.