To evaluate the function $f(x-1) = x^2 + 6x + 4$, we need to find the original function $f(x)$. To do this, we can substitute $u = x - 1$, so $x = u + 1$.

1. Substitute $x = u + 1$ into the given function: $f(u) = (u+1)^2 + 6(u+1) + 4$

2. Expand the expression: $f(u) = (u^2 + 2u + 1) + (6u + 6) + 4$

3. Simplify by combining like terms: $f(u) = u^2 + 2u + 1 + 6u + 6 + 4 = u^2 + 8u + 11$

Thus, the original function is: $f(x) = x^2 + 8x + 11$

Would you like more details or have any questions about this solution? Here are some related questions you might find interesting:

1. How would you evaluate $f(x+2)$ if $f(x) = x^2 + 8x + 11$?
2. Can you find the vertex of the parabola represented by $f(x) = x^2 + 8x + 11$?
3. What are the roots of the function $f(x) = x^2 + 8x + 11$?
4. How does the graph of $f(x) = x^2 + 8x + 11$ compare to the graph of $f(x) = x^2$?
5. What is the y-intercept of the function $f(x) = x^2 + 8x + 11$?
6. How would you determine the axis of symmetry for $f(x) = x^2 + 8x + 11$?
7. Can you express $f(x) = x^2 + 8x + 11$ in vertex form?
8. What is the minimum value of $f(x) = x^2 + 8x + 11$?

Tip: When dealing with functions, substituting and simplifying carefully is key to finding the correct form of the function.