1. **State the problem:** We have the function $f(x) = 14 - x^2$. (a) Find the slope of the tangent line to the graph $y = f(x)$ at $x = 2$ using the limit definition of the derivative. (b) Find the equation of the tangent line to the curve at $x = 2$. 2. **Recall the limit definition of the derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ This gives the slope of the tangent line at any point $x$. 3. **Apply the definition for $x=2$: ** $$\text{slope} = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$$ 4. **Calculate $f(2+h)$ and $f(2)$:** $$f(2+h) = 14 - (2+h)^2 = 14 - (4 + 4h + h^2) = 14 - 4 - 4h - h^2 = 10 - 4h - h^2$$ $$f(2) = 14 - 2^2 = 14 - 4 = 10$$ 5. **Substitute into the limit expression:** $$\lim_{h \to 0} \frac{(10 - 4h - h^2) - 10}{h} = \lim_{h \to 0} \frac{-4h - h^2}{h}$$ 6. **Simplify the fraction:** $$\frac{-4h - h^2}{h} = -4 - h$$ 7. **Evaluate the limit as $h \to 0$:** $$\lim_{h \to 0} (-4 - h) = -4$$ **Answer for (a):** The slope of the tangent line at $x=2$ is $-4$. 8. **Find the equation of the tangent line:** The tangent line at $x=2$ has slope $m = -4$ and passes through the point $(2, f(2)) = (2, 10)$. Use point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 10 = -4(x - 2)$$ Simplify: $$y - 10 = -4x + 8$$ $$y = -4x + 18$$ **Answer for (b):** The equation of the tangent line is $y = -4x + 18$.