## Calculus, 10th Edition (Anton)

$(a) \quad m_{tan} = 2x_0. \\ (b) \quad m_{tan} = -2.$
Let us find the slope of the tangent line to the graph of $f(x) = x^2 - 1$ at a general point $x = x_0$. \begin{align} m_{tan} &= \lim\limits_{h \to 0} \frac{f(x_0+h) - f(x_0)}{h} \\ &= \lim\limits_{h \to 0} \frac{(x_0+h)^2 - 1 - (x_0^2 - 1)}{h} \\ &= \lim\limits_{h \to 0} \frac{x_0^2 +2x_0h + h^2 - x_0^2}{h} \\ &= \lim\limits_{h \to 0} \frac{2x_0h + h^2}{h} \\ &= \lim\limits_{h \to 0} 2x_0 + h \\ &= 2x_0 \end{align} Now, let us find the the slope of the tangent line to the graph of $f$ at the point $x_0 = -1$. $$m_{tan} = 2x_0 = 2 \cdot (-1) = -2$$