The average rate of change of a function $f(x)$ over an interval $[a, b]$ is given by the formula:

$\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}.$

Here, $f(x) = x^2 - 4x$, $a = 5$, and $b = 5 + h$. Substituting into the formula:

$\text{Average Rate of Change} = \frac{f(5 + h) - f(5)}{(5 + h) - 5}.$

Step 1: Compute $f(5 + h)$

$f(5 + h) = (5 + h)^2 - 4(5 + h).$ Expanding: $(5 + h)^2 = 25 + 10h + h^2,$ $4(5 + h) = 20 + 4h,$ $f(5 + h) = (25 + 10h + h^2) - (20 + 4h) = h^2 + 6h + 5.$

Step 2: Compute $f(5)$

$f(5) = 5^2 - 4(5) = 25 - 20 = 5.$

Step 3: Substitute into the formula

$\text{Average Rate of Change} = \frac{f(5 + h) - f(5)}{(5 + h) - 5}.$ Substituting $f(5 + h) = h^2 + 6h + 5$ and $f(5) = 5$: $\text{Average Rate of Change} = \frac{(h^2 + 6h + 5) - 5}{h}.$

Simplify the numerator: $(h^2 + 6h + 5) - 5 = h^2 + 6h.$

$\text{Average Rate of Change} = \frac{h^2 + 6h}{h}.$

Step 4: Simplify the expression

Factor out $h$ in the numerator: $\text{Average Rate of Change} = \frac{h(h + 6)}{h}.$

For $h \neq 0$, cancel $h$: $\text{Average Rate of Change} = h + 6.$

Final Answer:

The average rate of change of $f(x) = x^2 - 4x$ over $[5, 5 + h]$ is: $\boxed{h + 6}.$

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Would you like further clarification or additional examples? 😊

Here are 5 related questions you may explore:

1. What is the significance of the average rate of change in real-world contexts?
2. How do you compute the instantaneous rate of change at a point?
3. What happens to the average rate of change as $h \to 0$?
4. How do you find the derivative of $f(x) = x^2 - 4x$?
5. What are some applications of the average rate of change in physics?

Tip: Always simplify algebraic expressions step-by-step to avoid errors when dealing with complex problems.