1. **Stating the problem:** We are given the function $f(x) = x^2 - 1$ and a point $x_0 = -1$. We need to (a) find a formula for the slope of the tangent line to the graph of $f$ at a general point $x = x_0$, and (b) use that formula to find the slope of the tangent line at $x_0 = -1$. 2. **Finding the slope formula (part a):** The slope of the tangent line to the curve $y = f(x)$ at $x = x_0$ is given by the derivative $f'(x_0)$. First, compute the derivative $f'(x)$ of $f(x) = x^2 - 1$. Apply the power rule for derivatives: $$f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(1) = 2x - 0 = 2x$$ So the formula for slope at any point $x = x_0$ is: $$\boxed{ f'(x_0) = 2x_0 }$$ 3. **Finding the slope at $x_0 = -1$ (part b):** Substitute $x_0 = -1$ into the slope formula: $$f'(-1) = 2 \times (-1) = -2$$ Therefore, the slope of the tangent line to the graph of $f$ at $x = -1$ is $-2$. **Final answer:** (a) The slope formula is $f'(x_0) = 2x_0$. (b) The slope at $x_0 = -1$ is $-2$.