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The problem asks to find and simplify the **difference quotient** for the function \( f(x) = 5x^2 - 3 \). The difference quotient formula is given by:

\[
\frac{f(x+h) - f(x)}{h}, \quad \text{where } h \neq 0.
\]

### Steps to solve:

1. **Find \( f(x+h) \)**:
   \[
   f(x+h) = 5(x+h)^2 - 3
   \]
   Expanding \( (x+h)^2 \):
   \[
   f(x+h) = 5(x^2 + 2xh + h^2) - 3 = 5x^2 + 10xh + 5h^2 - 3
   \]

2. **Substitute \( f(x+h) \) and \( f(x) \) into the difference quotient**:
   \[
   \frac{f(x+h) - f(x)}{h} = \frac{(5x^2 + 10xh + 5h^2 - 3) - (5x^2 - 3)}{h}
   \]
   Simplify the expression:
   \[
   = \frac{5x^2 + 10xh + 5h^2 - 3 - 5x^2 + 3}{h}
   \]
   \[
   = \frac{10xh + 5h^2}{h}
   \]

3. **Simplify the quotient**:
   Factor \( h \) from the numerator:
   \[
   = \frac{h(10x + 5h)}{h}
   \]
   Since \( h \neq 0 \), cancel \( h \) from the numerator and denominator:
   \[
   = 10x + 5h
   \]

Thus, the simplified form of the difference quotient is:

\[
10x + 5h
\]

Would you like further clarification or more details?

Here are 5 related questions to explore further:
1. How does the difference quotient relate to the derivative of a function?
2. What happens to the difference quotient as \( h \to 0 \)?
3. Can you apply the difference quotient to a linear function, like \( f(x) = 3x + 2 \)?
4. How does expanding \( (x + h)^2 \) work for higher degree polynomials?
5. Why is it important that \( h \neq 0 \) in the difference quotient?

**Tip:** The difference quotient is a foundational concept in calculus and leads directly to the definition of a derivative!

Find the difference quotient f(x+h) - f(x) / h for f(x) = 5x^2 - 3 and simplify.

Learn how to compute the difference quotient for the quadratic function f(x) = 5x^2 - 3. Follow a detailed step-by-step process to expand and simplify the expression. This is a key concept in calculus and suitable for students in Grades 9-12.

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