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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 \( a \)).

f(x) = 2x^2 + 7 for all values of x. What is the range of f(x)?

Learn how to find the range of the quadratic function f(x) = 2x^2 + 7. This example covers key algebraic concepts including the quadratic formula and the vertex theorem. Suitable for students in Grades 9-11.

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