1. **State the problem:** We have the curve $C$ defined by the equation $$y = x^3 - 10x^2 + 27x - 23$$ and a point $P(5, -13)$ on $C$. We want to find the equation of the tangent line $l$ at $P$, verify where $l$ meets $C$ again, and find the area of the region $R$ bounded by $C$ and $l$. 2. **Find the slope of the tangent line $l$ at $P$ by differentiation:** The derivative of $y$ with respect to $x$ is $$\frac{dy}{dx} = 3x^2 - 20x + 27$$ Evaluate at $x=5$: $$\frac{dy}{dx}\bigg|_{x=5} = 3(5)^2 - 20(5) + 27 = 75 - 100 + 27 = 2$$ So the slope $m$ of the tangent line at $P$ is $2$. 3. **Find the equation of the tangent line $l$:** Using point-slope form: $$y - y_1 = m(x - x_1)$$ where $(x_1, y_1) = (5, -13)$ and $m=2$: $$y - (-13) = 2(x - 5)$$ $$y + 13 = 2x - 10$$ $$y = 2x - 23$$ So the equation of $l$ is $$y = 2x - 23$$. 4. **Verify that $l$ meets $C$ again on the y-axis:** On the y-axis, $x=0$. Substitute into $l$: $$y = 2(0) - 23 = -23$$ Substitute $x=0$ into $C$: $$y = 0 - 0 + 0 - 23 = -23$$ Since both $C$ and $l$ have $y=-23$ at $x=0$, they meet again on the y-axis. 5. **Find the area of the region $R$ bounded by $C$ and $l$:** The area is given by the integral of the difference between the curve and the line over the interval from $x=0$ to $x=5$: $$\text{Area} = \int_0^5 \big[(x^3 - 10x^2 + 27x - 23) - (2x - 23)\big] \, dx$$ Simplify the integrand: $$x^3 - 10x^2 + 27x - 23 - 2x + 23 = x^3 - 10x^2 + 25x$$ So, $$\text{Area} = \int_0^5 (x^3 - 10x^2 + 25x) \, dx$$ Integrate term-by-term: $$\int x^3 \, dx = \frac{x^4}{4}, \quad \int -10x^2 \, dx = -\frac{10x^3}{3}, \quad \int 25x \, dx = \frac{25x^2}{2}$$ Evaluate from 0 to 5: $$\left[ \frac{x^4}{4} - \frac{10x^3}{3} + \frac{25x^2}{2} \right]_0^5 = \left( \frac{5^4}{4} - \frac{10 \times 5^3}{3} + \frac{25 \times 5^2}{2} \right) - 0$$ Calculate each term: $$\frac{5^4}{4} = \frac{625}{4}$$ $$\frac{10 \times 5^3}{3} = \frac{10 \times 125}{3} = \frac{1250}{3}$$ $$\frac{25 \times 5^2}{2} = \frac{25 \times 25}{2} = \frac{625}{2}$$ Combine: $$\text{Area} = \frac{625}{4} - \frac{1250}{3} + \frac{625}{2}$$ Find common denominator 12: $$\frac{625}{4} = \frac{1875}{12}, \quad \frac{1250}{3} = \frac{5000}{12}, \quad \frac{625}{2} = \frac{3750}{12}$$ So, $$\text{Area} = \frac{1875}{12} - \frac{5000}{12} + \frac{3750}{12} = \frac{1875 - 5000 + 3750}{12} = \frac{625}{12}$$ **Final answers:** (a) Equation of tangent line $l$ is $$y = 2x - 23$$ (b) $l$ meets $C$ again on the y-axis at $(0, -23)$. (c) The exact area of region $R$ is $$\frac{625}{12}$$.