To find $f(6)$ given that $f(x) = -3x^2 + 6x + 1$, we substitute $x = 6$ into the equation for $f(x)$.

$f(6) = -3(6)^2 + 6(6) + 1$

Now calculate each term:

$f(6) = -3(36) + 36 + 1$

$f(6) = -108 + 36 + 1$

$f(6) = -108 + 37$

$f(6) = -71$

So, $f(6) = -71$.

Would you like a detailed breakdown of any step, or have any questions?

Here are 5 related questions:

1. What is $f(0)$ for the same function?
2. Find the vertex of the parabola $f(x) = -3x^2 + 6x + 1$.
3. What are the roots of $f(x) = 0$?
4. How does the sign of the leading coefficient affect the graph of the quadratic function?
5. Find $f'(x)$ and evaluate $f'(6)$.

Tip: The sign of the leading coefficient in a quadratic function affects whether the parabola opens upwards or downwards!