The problem asks you to calculate the slope of secant lines for the function $f(x) = (x - 1)^2$ between various points $x_1$ and $x_2$. The formula for the slope of a secant line is given by:

$m_{\text{sec}} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

Step-by-step calculations:

The function is $f(x) = (x - 1)^2$, so first, we need to compute this function for each given $x$ value, then use the secant line formula.

1. For $x_1 = 1.0$ and $x_2 = 2.0$:

   • $f(1.0) = (1.0 - 1)^2 = 0$
   • $f(2.0) = (2.0 - 1)^2 = 1$

   Now, calculate the slope: $m_{\text{sec}} = \frac{1 - 0}{2.0 - 1.0} = 1$

2. For $x_1 = 1.0$ and $x_2 = 1.5$:

   • $f(1.0) = (1.0 - 1)^2 = 0$
   • $f(1.5) = (1.5 - 1)^2 = 0.25$

   Slope: $m_{\text{sec}} = \frac{0.25 - 0}{1.5 - 1.0} = 0.5$

3. For $x_1 = 1.0$ and $x_2 = 1.1$:

   • $f(1.0) = (1.0 - 1)^2 = 0$
   • $f(1.1) = (1.1 - 1)^2 = 0.01$

   Slope: $m_{\text{sec}} = \frac{0.01 - 0}{1.1 - 1.0} = 0.1$

4. For $x_1 = 1.0$ and $x_2 = 1.01$:

   • $f(1.0) = 0$
   • $f(1.01) = (1.01 - 1)^2 = 0.0001$

   Slope: $m_{\text{sec}} = \frac{0.0001 - 0}{1.01 - 1.0} = 0.01$

5. For $x_1 = 1.0$ and $x_2 = 1.001$:

   • $f(1.0) = 0$
   • $f(1.001) = (1.001 - 1)^2 = 0.000001$

   Slope: $m_{\text{sec}} = \frac{0.000001 - 0}{1.001 - 1.0} = 0.001$

These results give us the slope of the secant lines between the respective points.

Would you like more details on any part of this process? Also, here are some related questions:

1. How does the slope of the secant line change as $x_2$ approaches $x_1$?
2. What is the significance of calculating the secant slope for smaller intervals?
3. How is the secant slope related to the derivative of the function at a point?
4. Can you approximate the derivative of $f(x) = (x - 1)^2$ using these secant slopes?
5. How would the calculations change if the function were cubic instead of quadratic?

Tip: The slope of the secant line becomes closer to the derivative as $x_2$ approaches $x_1$.