The normal to the curve $y(x-2)(x-3)=x+6$ at the point where the curve intersects the $y$ -axis passes through the point:

1. Find the intersection with the $y$ -axis: Set $x=0$.
$$y(0-2)(0-3)=0+6⟹y(6)=6⟹y=1$$
Point is $P(0,1)$.2. Find the slope of the tangent at $(0,1)$:Rewrite the curve: $y=\frac{x+6}{x^2-5x+6}$ Differentiating:
$$\frac{dy}{dx}=\frac{(x^2-5x+6)(1)-(x+6)(2x-5)}{(x^2-5x+6{)}^2}$$
At $(0,1)$:
$$\frac{dy}{dx}=\frac{(6)(1)-(6)(-5)}{(6{)}^2}=\frac{6+30}{36}=\frac{36}{36}=1$$
3. Find the slope of the normal ($m_n$):
$$m_n=-\frac{1}{\text{slope of tangent}}=-1$$
4. Equation of the normal at $(0,1)$:
$$y-1=-1(x-0)⟹x+y-1=0$$
5. Check options for point $(1/2,1/2)$:
$$1/2+1/2-1=1-1=0$$
The point $(1/2,1/2)$ satisfies the equation.