We are given the quadratic function:

$f(x) = -x^2 - 4x - 5$

Now, we will calculate the values of $f(x)$ for each of the given $x$-values.

When $x = -5$:

$f(-5) = -(-5)^2 - 4(-5) - 5$ $f(-5) = -(25) + 20 - 5 = -25 + 20 - 5 = -10$

When $x = -3$:

$f(-3) = -(-3)^2 - 4(-3) - 5$ $f(-3) = -(9) + 12 - 5 = -9 + 12 - 5 = -2$

When $x = -1$:

$f(-1) = -(-1)^2 - 4(-1) - 5$ $f(-1) = -(1) + 4 - 5 = -1 + 4 - 5 = -2$

When $x = 1$:

$f(1) = -(1)^2 - 4(1) - 5$ $f(1) = -(1) - 4 - 5 = -1 - 4 - 5 = -10$

The completed table:

$x$	$f(x)$
$-5$	$-10$
$-3$	$-2$
$-1$	$-2$
$1$	$-10$

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Would you like further explanation or any clarification on the steps?

Follow-up Questions:

1. How do the values of $f(x)$ change as $x$ increases or decreases?
2. Can you identify the vertex of the quadratic function based on this table?
3. How would you graph this function based on the values in the table?
4. What would the equation of the axis of symmetry be for this quadratic function?
5. How would you find the x-intercepts for this function?

Tip:

To quickly find the axis of symmetry for any quadratic function $f(x) = ax^2 + bx + c$, use the formula:
$x = \frac{-b}{2a}$