Question

Find the equations of the tangent and the normal to the curve $y=\frac{x-7}{(x-2)(x-3)}$ at the point where it cuts the x-axis.

Answer 1

The equation cuts $\text{x-axis}$ at $(7,0)$

$\frac{dy}{dx}=\frac{(x^2-5x+6)-(x-7)(2x-5)}{(x^2-5x+6{)}^2}$

At $(7,0)$

Slope at tangent is $20$

Equation is $y=20(x-7)⟹20x-y-140=0$

Slope at Normal is $\frac{-1}{20}$

Equation is $y=\frac{-1}{20}(x-7)⟹x+20y-7=0$