We are solving two parts based on the given function $f(x) = -x^2 - 2x + 5$.

Part 1: $f(3 + a)$

Substitute $x = 3 + a$ into $f(x)$:

$f(3 + a) = -(3 + a)^2 - 2(3 + a) + 5$

Expand the terms:

$f(3 + a) = -(9 + 6a + a^2) - 2(3 + a) + 5$ $f(3 + a) = -9 - 6a - a^2 - 6 - 2a + 5$ $f(3 + a) = -a^2 - 8a - 10$

Part 2: $\frac{f(3 + a) - f(a)}{a}$

Substitute $x = a$ into $f(x)$:

$f(a) = -a^2 - 2a + 5$

Now calculate $f(3 + a) - f(a)$:

$f(3 + a) - f(a) = (-a^2 - 8a - 10) - (-a^2 - 2a + 5)$ $f(3 + a) - f(a) = -a^2 - 8a - 10 + a^2 + 2a - 5$ $f(3 + a) - f(a) = -6a - 15$

Divide by $a$:

$\frac{f(3 + a) - f(a)}{a} = \frac{-6a - 15}{a}$ $\frac{f(3 + a) - f(a)}{a} = -6 - \frac{15}{a}$

Final Answer:

The two results are:

1. $f(3 + a) = -a^2 - 8a - 10$
2. $\frac{f(3 + a) - f(a)}{a} = -6 - \frac{15}{a}$

The correct choice is:

$\boxed{(-a^2 - 8a - 10) \text{ and } (-6 - \frac{15}{a})}$

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Would you like more details or have any questions?

Related Questions:

1. How do we derive $f(x)$ if given as a quadratic polynomial?
2. Can you explain the meaning of $\frac{f(3 + a) - f(a)}{a}$ in terms of calculus?
3. What happens if $a = 0$ in the second part, and how do we handle it?
4. How do we simplify similar expressions for higher-degree polynomials?
5. Can we find a general formula for $f(x + a)$ given $f(x)$?

Tip: Always check your substitution carefully to avoid sign errors!