(c) Given $$ f(x)=x^{2}-3 x $$ (1) Determine the value of $$ \lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h} $$. (2) Explain what your answer in (1) means.

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$$ f(x) = x^2 - 3x $$

**Step 1. Evaluate $$ f(2+h) $$:**

$$
\begin{aligned}
f(2+h) & = (2+h)^2 - 3(2+h) \
       & = 4 + 4h + h^2 - 6 - 3h \
       & = h^2 + h - 2.
\end{aligned}
$$

**Step 2. Evaluate $$ f(2) $$:**

$$
f(2) = 2^2 - 3(2) = 4 - 6 = -2.
$$

**Step 3. Form the difference quotient:**

$$
\frac{f(2+h) - f(2)}{h} = \frac{(h^2 + h - 2) - (-2)}{h} = \frac{h^2 + h}{h}.
$$

**Step 4. Simplify the difference quotient:**

$$
\frac{h^2 + h}{h} = h + 1,\quad \text{(for } h \neq 0 \text{)}.
$$

**Step 5. Take the limit as $$ h \rightarrow 0 $$:**

$$
\lim_{h \to 0}(h + 1) = 1.
$$

**(1) Final Answer:**

$$
\lim_{h \to 0} \frac{f(2+h)-f(2)}{h} = 1.
$$

**(2) Explanation:**

The computed limit represents the derivative of $$ f(x) $$ at $$ x = 2 $$, denoted by $$ f'(2) $$. This derivative indicates the instantaneous rate of change of the function $$ f(x) $$ at that point, which is the slope of the tangent line to the curve $$ f(x) = x^2 - 3x $$ at $$ x = 2 $$. In other words, when $$ x = 2 $$, the function is increasing at a rate of $$ 1 $$ unit vertically for every unit increase in $$ x $$.
**(1) The limit is 1.** **(2) This means the slope of the tangent to the curve $$ f(x) = x^2 - 3x $$ at $$ x = 2 $$ is 1.

© Given
(1) Determine the value of .
(2) Explain what your answer in (1) means.

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© Given (1) Determine the value of . (2) Explain what your answer in (1) means.