The given function is $f(x) = 2x^2 + 7$.

This is a quadratic function, which opens upwards because the coefficient of $x^2$ is positive (2 in this case). Since it is a parabola opening upwards, its minimum value occurs at the vertex.

Finding the vertex:

The general form of a quadratic function is $f(x) = ax^2 + bx + c$. For your function, $a = 2$, $b = 0$, and $c = 7$.

The vertex occurs at $x = \frac{-b}{2a}$. Since $b = 0$, this simplifies to: $x = 0$

Substitute $x = 0$ into the function to find the minimum value of $f(x)$: $f(0) = 2(0)^2 + 7 = 7$

Thus, the minimum value of $f(x)$ is 7, and since the parabola opens upwards, the function has no maximum value, but it can take on all values greater than or equal to 7.

Range of the function:

The range of $f(x)$ is: $f(x) \geq 7$

So, the range is $[7, \infty)$.

Would you like further clarification or details on how to compute the vertex of a quadratic function?

Here are 5 related questions to extend this concept:

1. How do you determine the axis of symmetry for any quadratic function?
2. How can you find the maximum or minimum value of a general quadratic function?
3. What changes in the range if the quadratic coefficient $a$ is negative?
4. How does the constant $c$ in a quadratic function affect its graph?
5. What is the domain of any quadratic function?

Tip: For quadratic functions of the form $ax^2 + bx + c$, the vertex provides valuable information about the range, especially when combined with the direction the parabola opens (determined by $a$).