1. **Problem statement:** (a) Find the slope of the secant line through points (1,2) and (2,17) for the function $f(x) = 4x^2 + 3x - 5$. (b) Find the derivative $f'(x)$ using the limit definition and then find the equation of the tangent line at point (1,2). 2. **Formula for slope of secant line:** $$\text{slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$$ 3. **Calculate slope of secant line:** Given points $(1,2)$ and $(2,17)$, $$\text{slope} = \frac{17 - 2}{2 - 1} = \frac{15}{1} = 15$$ 4. **Limit definition of derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 5. **Calculate $f(x+h)$:** $$f(x+h) = 4(x+h)^2 + 3(x+h) - 5 = 4(x^2 + 2xh + h^2) + 3x + 3h - 5 = 4x^2 + 8xh + 4h^2 + 3x + 3h - 5$$ 6. **Calculate difference quotient:** $$\frac{f(x+h) - f(x)}{h} = \frac{(4x^2 + 8xh + 4h^2 + 3x + 3h - 5) - (4x^2 + 3x - 5)}{h} = \frac{8xh + 4h^2 + 3h}{h}$$ 7. **Cancel $h$ in numerator and denominator:** $$\frac{\cancel{h}(8x + 4h + 3)}{\cancel{h}} = 8x + 4h + 3$$ 8. **Take limit as $h \to 0$:** $$f'(x) = \lim_{h \to 0} (8x + 4h + 3) = 8x + 3$$ 9. **Find slope of tangent line at $x=1$:** $$f'(1) = 8(1) + 3 = 11$$ 10. **Equation of tangent line:** Using point-slope form $y - y_1 = m(x - x_1)$ with point $(1,2)$ and slope $11$: $$y - 2 = 11(x - 1)$$ $$y = 11x - 11 + 2 = 11x - 9$$ **Final answers:** (a) Slope of secant line = $15$ (b) Derivative $f'(x) = 8x + 3$, tangent line equation: $y = 11x - 9$