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Determine the slope of the secant line of f for each of the intervals indicated in the table

The function provided is \( f(x) = x^2 + 6x + 6 \). To find the slope of the secant line over the interval \([-2, x_2]\), we use the formula for the slope of the secant line:

\[
m_{\text{secant}} = \frac{f(x_2) - f(-2)}{x_2 - (-2)}
\]

where \( f(x) = x^2 + 6x + 6 \) and \( f(-2) \) needs to be calculated first.

### Step 1: Compute \( f(-2) \)
\[
f(-2) = (-2)^2 + 6(-2) + 6 = 4 - 12 + 6 = -2
\]

Now, using \( f(-2) = -2 \), we can calculate the secant slopes for each given \( x_2 \) in the table.

### Step 2: Calculate the slopes for each \( x_2 \)

1. For \( x_2 = -1 \):
   \[
   f(-1) = (-1)^2 + 6(-1) + 6 = 1 - 6 + 6 = 1
   \]
   Slope:
   \[
   m_{\text{secant}} = \frac{f(-1) - f(-2)}{-1 - (-2)} = \frac{1 - (-2)}{-1 + 2} = \frac{3}{1} = 3
   \]

2. For \( x_2 = -1.9 \):
   \[
   f(-1.9) = (-1.9)^2 + 6(-1.9) + 6 = 3.61 - 11.4 + 6 = -1.79
   \]
   Slope:
   \[
   m_{\text{secant}} = \frac{f(-1.9) - f(-2)}{-1.9 - (-2)} = \frac{-1.79 - (-2)}{-1.9 + 2} = \frac{0.21}{0.1} = 2.1
   \]

3. For \( x_2 = -1.99 \):
   \[
   f(-1.99) = (-1.99)^2 + 6(-1.99) + 6 = 3.9601 - 11.94 + 6 = -1.9799
   \]
   Slope:
   \[
   m_{\text{secant}} = \frac{f(-1.99) - f(-2)}{-1.99 - (-2)} = \frac{-1.9799 - (-2)}{-1.99 + 2} = \frac{0.0201}{0.01} = 2.01
   \]

4. For \( x_2 = -1.999 \):
   \[
   f(-1.999) = (-1.999)^2 + 6(-1.999) + 6 = 3.996001 - 11.994 + 6 = -1.997999
   \]
   Slope:
   \[
   m_{\text{secant}} = \frac{f(-1.999) - f(-2)}{-1.999 - (-2)} = \frac{-1.997999 - (-2)}{-1.999 + 2} = \frac{0.002001}{0.001} = 2.001
   \]

### Final Secant Slopes:
- For \( x_2 = -1 \), the slope is \( 3 \).
- For \( x_2 = -1.9 \), the slope is \( 2.1 \).
- For \( x_2 = -1.99 \), the slope is \( 2.01 \).
- For \( x_2 = -1.999 \), the slope is \( 2.001 \).

Let me know if you'd like more details or have any questions!

Here are some relative questions:
1. How is the secant line different from the tangent line?
2. What does the secant slope approach as \( x_2 \) approaches \(-2\)?
3. How can the concept of the derivative be explained through secant slopes?
4. What does the slope of the tangent line tell us about a function at a point?
5. How can secant lines help estimate instantaneous rates of change?

**Tip:** As \( x_2 \) gets closer to \( -2 \), the secant slope approaches the derivative at \( x = -2 \), which gives the slope of the tangent line.

The function f(x) = x^2 + 6x + 6 is graphed below. Determine the slope of the secant line of f for each of the intervals indicated in the table.

Learn how to find the slope of the secant line for the quadratic function f(x) = x^2 + 6x + 6 over different intervals. This math problem covers essential concepts in algebra and calculus, including quadratic functions and secant line slopes. Suitable for high school students in Grades 10-12.

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