Question

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

Answer 1

Note that on x- axis, $y=0$. So the equation of the curve, When $y=0$, gives $x=7$. Thus, the curve cuts the x-axis at $(7,0)$, Now differentiating the equation of curve with respect to x, we obtain
$\frac{dy}{dx}=\frac{1-y(2x-5)}{(x-2)(x-3)}$ (Why?)
${\frac{dy}{dx}]}_{7,0}=\frac{1-0}{\left(5\right)\left(4\right)}=\frac{1}{20}$