2. Evaluate and simplify the difference quotient \( \frac{f(x+h)-f(x)}{h} \) for the function \( f(x)=x^{2}-4 \).

To evaluate and simplify the difference quotient \( \frac{f(x+h)-f(x)}{h} \) for the function \( f(x) = x^2 - 4 \), we first need to compute \( f(x+h) \). 1. **Calculate \( f(x+h) \)**: \[ f(x+h) = (x+h)^2 - 4 \] Expanding \( (x+h)^2 \): \[ (x+h)^2 = x^2 + 2xh + h^2 \] Therefore, \[ f(x+h) = x^2 + 2xh + h^2 - 4 \] 2. **Calculate \( f(x+h) - f(x) \)**: We already have \( f(x) = x^2 - 4 \). Now we find the difference: \[ f(x+h) - f(x) = (x^2 + 2xh + h^2 - 4) - (x^2 - 4) \] Simplifying this expression: \[ f(x+h) - f(x) = x^2 + 2xh + h^2 - 4 - x^2 + 4 \] The \( x^2 \) and \(-4\) terms cancel out: \[ f(x+h) - f(x) = 2xh + h^2 \] 3. **Substitute into the difference quotient**: Now we substitute \( f(x+h) - f(x) \) into the difference quotient: \[ \frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2}{h} \] 4. **Simplify the expression**: We can factor \( h \) out of the numerator: \[ \frac{2xh + h^2}{h} = \frac{h(2x + h)}{h} \] As long as \( h \neq 0 \), we can cancel \( h \): \[ 2x + h \] Thus, the simplified form of the difference quotient is: \[ \boxed{2x + h} \]