1. Let's clarify the problem: You want to find the domain and vertex of a function, typically a quadratic function. 2. The domain of a quadratic function $f(x) = ax^2 + bx + c$ is all real numbers, because you can plug any real number into $x$ and get a valid output. 3. The vertex of a quadratic function is the point where the function reaches its maximum or minimum value. 4. The formula for the vertex $x$-coordinate is $x = -\frac{b}{2a}$. 5. To find the $y$-coordinate of the vertex, substitute $x$ back into the function: $y = f\left(-\frac{b}{2a}\right)$. 6. Example: For $f(x) = 2x^2 - 4x + 1$, domain is all real numbers. 7. Calculate vertex $x$-coordinate: $x = -\frac{-4}{2 \times 2} = \frac{4}{4} = 1$. 8. Calculate vertex $y$-coordinate: $y = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1$. 9. So, vertex is at $(1, -1)$. 10. Summary: Domain is $(-\infty, \infty)$ and vertex is at $(1, -1)$ for this example.