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### Let's break down the problem step by step:

#### a. **Graph the function:**
The given function is \( f(x) = x^2 - 10x + 21 \).

- The correct graph among the options is **C**, as it shows the parabola correctly based on the function, which opens upwards. You correctly chose this.

#### b. **Identify the point where the function has a tangent line with zero slope:**
To find the point where the function has a tangent line with zero slope, we need to compute the derivative of the function and set it equal to zero.

1. The derivative of \( f(x) \) is:
   \[
   f'(x) = 2x - 10
   \]
2. Set the derivative equal to zero to find where the slope is zero:
   \[
   2x - 10 = 0
   \]
   Solving this gives \( x = 5 \).
3. To find the corresponding \( y \)-coordinate, substitute \( x = 5 \) back into the original function:
   \[
   f(5) = 5^2 - 10(5) + 21 = 25 - 50 + 21 = -4
   \]
   So the point is \( (5, -4) \), which you have already correctly identified.

#### c. **Confirm by making a table of slopes of secant lines:**
Now, let’s compute the slopes of the secant lines. The table shows two intervals, and we will use the formula for the slope of a secant line between two points \( (x_1, f(x_1)) \) and \( (x_2, f(x_2)) \), which is:

\[
\text{slope of secant line} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}
\]

For the point \( a = 5 \) (where the slope is zero), let's complete the table:

1. **Interval [a, a+0.5]:**  
   You already have this: the slope is 0.5.
   
2. **Interval [a, a+0.1]:**  
   For \( x_2 = 5 + 0.1 = 5.1 \),
   \[
   f(5.1) = (5.1)^2 - 10(5.1) + 21 = 26.01 - 51 + 21 = -3.99
   \]
   Now, the slope of the secant line is:
   \[
   \frac{-3.99 - (-4)}{5.1 - 5} = \frac{-3.99 + 4}{0.1} = \frac{0.01}{0.1} = 0.1
   \]

Thus, the slope for the second interval is **0.1**.

---

### Final Answer for the table:
| Interval      | Slope of the secant line |
|---------------|--------------------------|
| [a, a + 0.5]  | 0.5                      |
| [a, a + 0.1]  | 0.1                      |

Would you like more details or have any other questions?

---

### Five Related Questions:
1. What is the geometric meaning of a secant line in relation to the tangent line?
2. How do changes in the interval size affect the slope of the secant line as it approaches the tangent point?
3. How would you compute the tangent line's equation at \( x = 5 \)?
4. Can you find another point where the function has a horizontal tangent line (if any)?
5. What is the importance of the second derivative in understanding concavity near \( x = 5 \)?

### Tip:
When calculating secant slopes, always remember that smaller intervals will give you a better approximation of the tangent slope at the point.

Graph the function f(x) = x^2 - 10x + 21. Identify the point where the function has a tangent line with zero slope. Calculate secant slopes for intervals around that point.

Learn how to graph the quadratic function f(x) = x^2 - 10x + 21, find points with a tangent line of zero slope, and calculate secant slopes for different intervals. This problem introduces key concepts in calculus and algebra, suitable for high school students.

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