To find $k(a + 3)$ for the given function $k(x) = -x^2 - 3x + 4$, we'll substitute $x = a + 3$ into the function.

Given:

$k(x) = -x^2 - 3x + 4$

Substitute $x = a + 3$:

$k(a + 3) = -(a + 3)^2 - 3(a + 3) + 4$

Now, expand and simplify:

1. Expand $(a + 3)^2$: $(a + 3)^2 = a^2 + 6a + 9$

2. Multiply by $-1$: $-(a^2 + 6a + 9) = -a^2 - 6a - 9$

3. Expand and simplify $-3(a + 3)$: $-3(a + 3) = -3a - 9$

4. Add the constant term $+4$:

So,

$k(a + 3) = -a^2 - 6a - 9 - 3a - 9 + 4$

Combine like terms:

$k(a + 3) = -a^2 - 9a - 14$

Thus, the final expression for $k(a + 3)$ is:

$k(a + 3) = -a^2 - 9a - 14$

Would you like more details on this solution or have any questions?

Here are 5 related questions to consider:

1. What is the vertex of the function $k(x) = -x^2 - 3x + 4$?
2. How does the value of $k(a + 3)$ change if $a = -2$?
3. What are the roots of the equation $k(x) = 0$?
4. How does the graph of $k(x)$ compare to the graph of $x^2 + 3x - 4$?
5. What is the maximum value of $k(x)$ for real $x$?

Tip: Always double-check your algebraic expansions and simplifications to ensure accuracy in calculations.