https://file.aiquickdraw.com/mathbot/file/17265541317715u6546iw.jpg

Let's break down the problems one by one based on the image you've provided:

### 1. **Difference Quotient**:

We are given \( f(x) = 5 - x^2 \) and asked to calculate the difference quotient \( \frac{f(x) - f(5)}{x - 5} \).

#### Step 1: Find \( f(5) \)
\( f(5) = 5 - (5)^2 = 5 - 25 = -20 \).

#### Step 2: Set up the difference quotient
\[
\frac{f(x) - f(5)}{x - 5} = \frac{(5 - x^2) - (-20)}{x - 5} = \frac{5 - x^2 + 20}{x - 5} = \frac{25 - x^2}{x - 5}.
\]

#### Step 3: Simplify
Notice that \( 25 - x^2 \) can be factored as \( (5 - x)(5 + x) \). Thus:
\[
\frac{25 - x^2}{x - 5} = \frac{(5 - x)(5 + x)}{x - 5}.
\]
Now, since \( 5 - x = -(x - 5) \), we can rewrite the expression as:
\[
\frac{-(x - 5)(5 + x)}{x - 5}.
\]
The \( x - 5 \) terms cancel out, leaving:
\[
-(5 + x).
\]
Thus, the difference quotient is \( -(5 + x) \).

### 2. **Derivative \( f'(5) \) using the limit definition**:

The derivative at \( x = 5 \) is found by taking the limit:
\[
f'(5) = \lim_{x \to 5} \frac{f(x) - f(5)}{x - 5}.
\]

We already know that \( \frac{f(x) - f(5)}{x - 5} = -(5 + x) \). Taking the limit as \( x \to 5 \):
\[
f'(5) = -(5 + 5) = -10.
\]
So, \( f'(5) = -10 \).

### 3. **Equation of the Tangent Line**:

The equation of the tangent line to \( f(x) \) at \( x = 5 \) is given by the point-slope form:
\[
y - f(5) = f'(5)(x - 5).
\]
We know that \( f(5) = -20 \) and \( f'(5) = -10 \), so the equation becomes:
\[
y - (-20) = -10(x - 5),
\]
or equivalently:
\[
y + 20 = -10(x - 5).
\]
Simplifying this:
\[
y = -10(x - 5) - 20.
\]
Expanding the right-hand side:
\[
y = -10x + 50 - 20 = -10x + 30.
\]
Thus, the equation of the tangent line is \( y = -10x + 30 \).

---

### Summary:
1. Difference Quotient: \( -(5 + x) \).
2. Derivative \( f'(5) = -10 \).
3. Equation of Tangent Line: \( y = -10x + 30 \).

---

Would you like more details on any of these steps? Here are five related questions to explore further:

1. What is the geometric interpretation of the derivative at a point?
2. How do we generalize the difference quotient for any function?
3. How can we find the second derivative of \( f(x) = 5 - x^2 \)?
4. What does the slope of the tangent line represent in real-world applications?
5. How does the symmetry of \( f(x) = 5 - x^2 \) affect its derivative and tangent lines?

**Tip**: When simplifying difference quotients, always check if the numerator can be factored to cancel terms in the denominator.

Calculate the difference quotient for f(x) = 5 - x^2 at a = 5. Then find f'(5) using the limit definition and determine the equation of the tangent line at a = 5.

Learn how to calculate the difference quotient and derivative for f(x) = 5 - x^2 at x = 5 using the limit definition. Explore the equation of the tangent line at this point with step-by-step explanations, factoring techniques, and calculus fundamentals.

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