The function you have provided is:

$f(x) = 2x^2$

This is a quadratic function, where $a = 2$, and it has the general form $f(x) = ax^2 + bx + c$, with $b = 0$ and $c = 0$. Let's go over some key properties of this function:

1. Vertex: Since there is no linear term ($bx$) or constant term ($c$), the vertex is at the origin, $(0,0)$.

2. Axis of symmetry: The axis of symmetry is the vertical line $x = 0$, which passes through the vertex.

3. Direction of the parabola: The coefficient of $x^2$ is positive, so the parabola opens upward.

4. Domain: The function is defined for all real values of $x$, so the domain is $(-\infty, \infty)$.

5. Range: Since the parabola opens upward and the minimum value occurs at $f(0) = 0$, the range is $[0, \infty)$.

6. Growth rate: As $x$ increases or decreases away from 0, the value of $f(x)$ grows quadratically, meaning it grows faster than a linear function.

Would you like more details or any specific questions answered?

Here are 5 related questions:

1. How would the graph change if the coefficient of $x^2$ was negative?
2. What are the intercepts of this quadratic function?
3. How can the derivative of this function help describe its behavior?
4. How does this quadratic compare to $f(x) = x^2$?
5. Can you express this function in vertex form, even though the vertex is already at the origin?

Tip: The coefficient of $x^2$ controls how "wide" or "narrow" the parabola is—larger values make the graph narrower.