Let's solve the problem step by step.

Given Function:

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

1. Domain (D):

The function $f(x) = (x + 4)^2 - 5$ is a quadratic function. Since there are no restrictions (like division by zero or square roots of negative numbers), the domain of any quadratic function is all real numbers.

Domain: $D: (-\infty, \infty)$

2. Range (R):

To find the range, we analyze the vertex form of the quadratic function. The function is written as: $f(x) = (x + 4)^2 - 5$

This is a parabola that opens upwards (because the coefficient of the quadratic term is positive). The vertex occurs at $x = -4$, and the minimum value of the function occurs at the vertex. Substituting $x = -4$ into the function: $f(-4) = (-4 + 4)^2 - 5 = 0 - 5 = -5$ Thus, the minimum value of $f(x)$ is $-5$. Since the parabola opens upwards, the function's values will increase from $-5$ to infinity.

Range: $R: [-5, \infty)$

Would you like more details on this solution? Here are 5 related questions to expand your understanding:

1. How do you convert a general quadratic function to vertex form?
2. Why does the range of the function start from the minimum value of $-5$?
3. What would change in the domain or range if the quadratic function opened downwards?
4. Can the range of a quadratic function ever include negative infinity?
5. How would transformations affect the domain and range of a quadratic function?

Tip: The domain of most polynomial functions (like quadratics) is always all real numbers unless specific restrictions are involved, such as square roots or division by zero.