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.