The function given is:

$f(x) = 2x^2 + 5\sqrt{x - 2}$

To determine the domain of $f(x)$, we need to find all the values of $x$ for which the function is defined.

Step 1: Analyze the quadratic term

The term $2x^2$ is defined for all real numbers. There are no restrictions on $x$ here.

Step 2: Analyze the square root term

The square root function $\sqrt{x - 2}$ is only defined when the expression inside the square root is non-negative. Therefore, we need:

$x - 2 \geq 0$

Solving this inequality:

$x \geq 2$

Conclusion:

The domain of $f(x)$ is all real numbers $x$ such that $x \geq 2$.

So, the complete statement would be:

"The domain for $f(x)$ is all real numbers greater than or equal to 2."

Would you like any further details or clarifications?

Here are 5 related questions you might find helpful:

1. How would the domain change if the square root term was $\sqrt{3 - x}$?
2. What is the range of the function $f(x) = 2x^2 + 5\sqrt{x - 2}$?
3. How do you find the domain of a function involving both square roots and fractions?
4. Can you graph the function $f(x) = 2x^2 + 5\sqrt{x - 2}$ and interpret its behavior?
5. How does the coefficient of the square root, 5, affect the shape of the graph?

Tip: Always ensure that the expression inside a square root is non-negative when finding the domain of functions involving square roots.