$$\begin{align} y &= {x^2-4x+3 \over \sqrt{x-1}} \\[2ex] {dy \over dx} &= { (x-1)^\frac{1}{2} \dfrac{dx^2-4x+3 }{dx} - (x^2-4x+3) \dfrac{d (x-1)^\frac{1}{2}}{dx} \over x-1} \\[3ex] &= {(x-1)^\frac{1}{2} (2x-4) - (x^2-4x+3) {\dfrac{1}{2} (x-1)^\frac{-1}{2}} \over x-1} \\[3ex] &= {2(x-1)^\frac{1}{2} (x-2) - \dfrac{1}{2}(x^2-4x+3) {(x-1)^\frac{-1}{2}} \over x-1} \\[2ex] &= {2(x-1)^\frac{1}{2} (x-2) - \dfrac{1}{2}(x^2-4x+3) {(x-1)^\frac{-1}{2}} \over x-1} \cdot \dfrac{2(x-1)^\frac{1}{2}}{2(x-1)^\frac{1}{2}}\\[2ex] &= { 4(x-1)(x-2) - (x^2-4x+3) \over 2(x-1)^\frac{3}{2} }\\[2ex] &= { 4x^2-12x+8 - x^2 + 4x -3 \over 2(x-1)^\frac{3}{2} }\\[2ex] &= { 3x^2-8x+5 \over 2(x-1)^\frac{3}{2}}\\[2ex] \end{align}$$

$$y = x^3-4x+3 \\[2ex] \begin{align} \left. y \right|_{x=1} &= (1)^3 -4(1)+3 \\[2ex] &= -2 \\[2ex] \end{align}$$

$$\begin{align} \dfrac{dy}{dx} &= 3x^2 - 4 \\[2ex] \left.\dfrac{dy}{dx} \right|_{x=1} &= 3(1)^2 - 4 \\[2ex] &= -1 \end{align}$$

$$\\ \text{設斜率方程為} \ y=mx+c \\[2ex] \begin{align} \therefore m &= -1\\[2ex] y &= -x + c \\[2ex] \end{align}$$

$$\\ \text{代} \ (1, -2) \text{入} \ y=-x+c \\[2ex] \begin{align} -2 &= -(1) + c\\[2ex] c &= -1 \\[2ex] \therefore y&=-x-1 \end{align}$$

$$y = x^3-4x+3 \\[2ex] \begin{align} \left. y \right|_{x=2} &= (2)^3 -4(2)+3 \\[2ex] &= 3 \\[2ex] \end{align}$$

$$\begin{align} \dfrac{dy}{dx} &= 3x^2 - 4 \\[2ex] \left.\dfrac{dy}{dx} \right|_{x=2} &= 3(2)^2 - 4 \\[2ex] &= 8 \end{align}$$

$$\\ \text{設斜率方程為} \ y=mx+c \\[2ex] \begin{align} \therefore m &= \frac{-1}{8}\\[2ex] y &= \frac{-x}{8} + c \\[2ex] \end{align}$$

$$\\ \text{代} \ (2, 3) \text{入} \ y = \frac{-x}{8} + c \\[2ex] \begin{align} 3 &= \frac{-2}{8} + c\\[2ex] c &= \frac{13}{4} \\[2ex] \end{align}$$

$$\therefore y=\frac{-x}{8} + \frac{13}{4} \\[2ex] \therefore x+8y =26 \\[2ex]$$