1. **Problem statement:** Given the function $f(x) = x^2 - 2x$, we need to find: (A) The slope of the secant line joining points $(2, f(2))$ and $(7, f(7))$. (B) The slope of the secant line joining points $(5, f(5))$ and $(5+h, f(5+h))$. (C) The slope of the tangent line at the point $(5, f(5))$. (D) The equation of the tangent line at the point $(5, f(5))$. --- 2. **Formula for slope of secant line:** $$\text{slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$$ 3. **Formula for slope of tangent line:** The slope of the tangent line at $x=a$ is the derivative $f'(a)$, which can be found by: $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ 4. **Calculate $f(x)$ values:** $$f(2) = 2^2 - 2 \times 2 = 4 - 4 = 0$$ $$f(7) = 7^2 - 2 \times 7 = 49 - 14 = 35$$ $$f(5) = 5^2 - 2 \times 5 = 25 - 10 = 15$$ 5. **(A) Slope of secant line between $(2, f(2))$ and $(7, f(7))$:** $$\text{slope} = \frac{f(7) - f(2)}{7 - 2} = \frac{35 - 0}{5} = \frac{35}{5} = 7$$ 6. **(B) Slope of secant line between $(5, f(5))$ and $(5+h, f(5+h))$:** Calculate $f(5+h)$: $$f(5+h) = (5+h)^2 - 2(5+h) = (25 + 10h + h^2) - (10 + 2h) = 25 + 10h + h^2 - 10 - 2h = 15 + 8h + h^2$$ Slope: $$\frac{f(5+h) - f(5)}{(5+h) - 5} = \frac{(15 + 8h + h^2) - 15}{h} = \frac{8h + h^2}{h}$$ Cancel $h$: $$\frac{\cancel{h}(8 + h)}{\cancel{h}} = 8 + h$$ 7. **(C) Slope of tangent line at $x=5$:** Take the limit as $h \to 0$: $$f'(5) = \lim_{h \to 0} (8 + h) = 8$$ 8. **(D) Equation of tangent line at $(5, f(5))$:** Use point-slope form: $$y - y_1 = m(x - x_1)$$ Where $m = 8$, $x_1 = 5$, $y_1 = 15$: $$y - 15 = 8(x - 5)$$ Simplify: $$y - 15 = 8x - 40$$ $$y = 8x - 40 + 15$$ $$y = 8x - 25$$ **Final answers:** (A) Slope of secant line = $7$ (B) Slope of secant line = $8 + h$ (C) Slope of tangent line = $8$ (D) Equation of tangent line: $y = 8x - 25$