To find the slope of the secant line through the points where $x = -3$ and $x = 1$ for the function $f(x) = x^2 + x$, follow these steps:

1. Calculate $f(-3)$: $f(-3) = (-3)^2 + (-3) = 9 - 3 = 6$

2. Calculate $f(1)$: $f(1) = (1)^2 + 1 = 1 + 1 = 2$

3. Find the slope of the secant line: The formula for the slope of the secant line between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $m = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

   Substituting the values we found: $m = \frac{f(1) - f(-3)}{1 - (-3)} = \frac{2 - 6}{1 + 3} = \frac{-4}{4} = -1$

So, the slope of the secant line is $m = -1$.

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Related Questions:

1. What is the equation of the secant line between these two points?
2. How do you find the derivative of the function $f(x) = x^2 + x$?
3. What is the slope of the tangent line to the function $f(x) = x^2 + x$ at $x = 1$?
4. How do you find the average rate of change of the function over the interval $[-3, 1]$?
5. What is the y-intercept of the secant line between these two points?

Tip: The slope of a secant line represents the average rate of change of the function over a specified interval.