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!