Let's solve the problem step by step.

Function:

The given function is:
$f(x) = (x - 2)^3 + 5$

1. Domain

The domain is already specified as "all real $x$" because $f(x)$ is a polynomial.

2. $y$-Intercept

To find the $y$-intercept, set $x = 0$:
$f(0) = (0 - 2)^3 + 5 = (-2)^3 + 5 = -8 + 5 = -3$
So, the $y$-intercept is $-3$.

3. Derivative

The derivative of $f(x) = (x - 2)^3 + 5$ is: $f'(x) = 3(x - 2)^2$

In expanded form: $f'(x) = 3x^2 - 12x + 12$

So the derivative is $3x^2 - 12x + 12$.

4. Number of Extrema

To find extrema, we solve $f'(x) = 0$:
$3(x - 2)^2 = 0 \implies x - 2 = 0 \implies x = 2$
The derivative equals zero at only one point $x = 2$, but this is not an extremum since the derivative does not change sign (it's a point of inflection). Hence, the number of extrema is 0.

5. Second Derivative

The second derivative is: $f''(x) = 6(x - 2)$

In expanded form: $f''(x) = 6x - 12$

6. Inflection Point

An inflection point occurs where $f''(x) = 0$:
$6(x - 2) = 0 \implies x = 2$
At $x = 2$, substitute into $f(x)$ to find $y$:
$f(2) = (2 - 2)^3 + 5 = 0 + 5 = 5$
Thus, the inflection point is $(2, 5)$.

Final Answers:

1. $y$-Intercept: $-3$
2. Derivative: $3x^2 - 12x + 12$
3. Number of Extrema: $0$
4. Second Derivative: $6x - 12$
5. Inflection Point: $(2, 5)$

Would you like further clarification?

Related Questions:

1. What are the conditions for extrema in polynomial functions?
2. Why does $f'(x) = 0$ at $x = 2$ not result in an extremum?
3. Can a cubic polynomial have multiple inflection points?
4. How can you determine concavity from the second derivative?
5. What happens to the graph of $f(x)$ as $x \to \infty$ and $x \to -\infty$?

Tip: For higher-order polynomials, always check sign changes in derivatives to confirm extrema or inflection points.