We are asked to find the slope of the secant line for the function $f(x) = 2x^2 - 12x + 10$ over the interval $[4, x_2]$ for various values of $x_2$ as indicated in the table.

The formula for the slope of the secant line between two points $x_1$ and $x_2$ on the function $f(x)$ is given by:

$\text{Secant Slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

In this case, $x_1 = 4$ and we will calculate the secant slope for each $x_2$ given in the table: 4.001, 4.01, 4.1, and 5.

First, let’s calculate $f(x)$ at the points $x_1 = 4$ and $x_2$ values.

1. Calculate $f(4)$:

$f(4) = 2(4)^2 - 12(4) + 10 = 2(16) - 48 + 10 = 32 - 48 + 10 = -6$

Now, we will calculate $f(x_2)$ for each $x_2$ value and then use the secant slope formula:

1. For $x_2 = 4.001$:

$f(4.001) = 2(4.001)^2 - 12(4.001) + 10 = 2(16.008001) - 48.012 + 10 = 32.016002 - 48.012 + 10 = -5.995998$

Secant slope: $\frac{f(4.001) - f(4)}{4.001 - 4} = \frac{-5.995998 - (-6)}{0.001} = \frac{0.004002}{0.001} = 4.002$

2. For $x_2 = 4.01$:

$f(4.01) = 2(4.01)^2 - 12(4.01) + 10 = 2(16.0801) - 48.12 + 10 = 32.1602 - 48.12 + 10 = -5.9598$

Secant slope: $\frac{f(4.01) - f(4)}{4.01 - 4} = \frac{-5.9598 - (-6)}{0.01} = \frac{0.0402}{0.01} = 4.02$

3. For $x_2 = 4.1$:

$f(4.1) = 2(4.1)^2 - 12(4.1) + 10 = 2(16.81) - 49.2 + 10 = 33.62 - 49.2 + 10 = -5.58$

Secant slope: $\frac{f(4.1) - f(4)}{4.1 - 4} = \frac{-5.58 - (-6)}{0.1} = \frac{0.42}{0.1} = 4.2$

4. For $x_2 = 5$:

$f(5) = 2(5)^2 - 12(5) + 10 = 2(25) - 60 + 10 = 50 - 60 + 10 = 0$

Secant slope: $\frac{f(5) - f(4)}{5 - 4} = \frac{0 - (-6)}{1} = \frac{6}{1} = 6$

Table of Secant Slopes:

Do you have any questions or need more details?

Here are some relative questions for practice:

1. How would you compute the secant slope if the function was $f(x) = 3x^2 + 5x - 7$?
2. What happens to the secant slope as $x_2$ approaches $x_1$?
3. Can you calculate the secant slope if $x_1 = 3$ and $x_2 = 6$ for the function $f(x) = x^2 - x$?
4. What is the relationship between secant slope and tangent slope?
5. How does the secant line change as $x_2$ becomes further from $x_1$?

Tip: The secant line slope approaches the tangent slope as the interval between the two points gets smaller.