what kind of function can you give an example 

Sure, let's consider a classic example of a function in calculus: the quadratic function.

### Example: Quadratic Function

A quadratic function is a polynomial function of degree 2, and it can be written in the form:

\[ f(x) = ax^2 + bx + c \]

where \( a \), \( b \), and \( c \) are constants, and \( a \neq 0 \).

#### Let's work with a specific example:

\[ f(x) = 2x^2 - 3x + 1 \]

### Key Calculus Concepts Applied to this Function

1. **Finding the Derivative (First Derivative):**
   - The first derivative of a function gives us the slope of the tangent line at any point \( x \).
   - For \( f(x) = 2x^2 - 3x + 1 \):
     \[ f'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(3x) + \frac{d}{dx}(1) \]
     \[ f'(x) = 4x - 3 \]

2. **Critical Points:**
   - Critical points occur where the first derivative is zero or undefined.
   - Set \( f'(x) = 0 \):
     \[ 4x - 3 = 0 \]
     \[ x = \frac{3}{4} \]

3. **Second Derivative:**
   - The second derivative provides information about the concavity of the function.
   - For \( f(x) = 2x^2 - 3x + 1 \):
     \[ f''(x) = \frac{d}{dx}(4x - 3) \]
     \[ f''(x) = 4 \]

   - Since \( f''(x) = 4 > 0 \), the function is concave up, and the critical point at \( x = \frac{3}{4} \) is a local minimum.

4. **Vertex of the Parabola:**
   - The vertex form of a quadratic function is another useful representation.
   - The vertex can be found using the formula \( x = -\frac{b}{2a} \):
     \[ x = -\frac{-3}{2 \cdot 2} \]
     \[ x = \frac{3}{4} \]

   - Substitute \( x = \frac{3}{4} \) back into \( f(x) \) to find the y-coordinate of the vertex:
     \[ f\left(\frac{3}{4}\right) = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + 1 \]
     \[ f\left(\frac{3}{4}\right) = 2 \cdot \frac{9}{16} - 3 \cdot \frac{3}{4} + 1 \]
     \[ f\left(\frac{3}{4}\right) = \frac{18}{16} - \frac{9}{4} + 1 \]
     \[ f\left(\frac{3}{4}\right) = \frac{18}{16} - \frac{36}{16} + \frac{16}{16} \]
     \[ f\left(\frac{3}{4}\right) = \frac{18 - 36 + 16}{16} \]
     \[ f\left(\frac{3}{4}\right) = \frac{-2}{16} \]
     \[ f\left(\frac{3}{4}\right) = -\frac{1}{8} \]

   - So, the vertex of the function \( f(x) = 2x^2 - 3x + 1 \) is at \( \left(\frac{3}{4}, -\frac{1}{8}\right) \).

### Summary:
The function \( f(x) = 2x^2 - 3x + 1 \) has:
- A derivative \( f'(x) = 4x - 3 \)
- A critical point at \( x = \frac{3}{4} \), which is a local minimum
- A vertex at \( \left(\frac{3}{4}, -\frac{1}{8}\right) \)

Would you like more details or have any questions? Here are some related questions to further explore:

1. How do you find the derivative of a polynomial function?
2. What is the significance of the second derivative in calculus?
3. Can you explain the process of completing the square for a quadratic function?
4. How do you determine if a critical point is a maximum or minimum?
5. What are inflection points and how do you find them?
6. How do you graph a quadratic function using its vertex and intercepts?
7. Can you provide an example of a real-world application of quadratic functions?
8. How do you solve a quadratic equation using the quadratic formula?

**Tip:** Understanding the graphical representation of a function alongside its algebraic form can significantly enhance your comprehension of calculus concepts.

Explore a detailed example of a quadratic function, f(x) = 2x^2 - 3x + 1, with comprehensive calculus analysis. Learn about derivatives, critical points, and the vertex of the parabola. Suitable for advanced high school students or anyone studying calculus concepts.

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