The normal to the curve y(x - 2)(x - 3) = x + | vedclass

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(C) The equation of the curve is $y(x - 2)(x - 3) = x + 6$,which can be written as $y = \frac{x + 6}{x^2 - 5x + 6}$.At the $y$-axis,$x = 0$. Substitut

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$(\frac{1}{2}, \frac{1}{2})$

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(C) The equation of the curve is $y(x - 2)(x - 3) = x + 6$,which can be written as $y = \frac{x + 6}{x^2 - 5x + 6}$.At the $y$-axis,$x = 0$. Substituting $x = 0$ into the equation,we get $y(0 - 2)(0 - 3) = 0 + 6$,which simplifies to $y(6) = 6$,so $y = 1$. Thus,the point of intersection is $(0, 1)$.Now,differentiate $y = \frac{x + 6}{x^2 - 5x + 6}$ with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{(x^2 - 5x + 6)(1) - (x + 6)(2x - 5)}{(x^2 - 5x + 6)^2}$.At the point $(0, 1)$,we have $x = 0$ and $x^2 - 5x + 6 = 6$:$\frac{dy}{dx} = \frac{(6)(1) - (6)(-5)}{(6)^2} = \frac{6 + 30}{36} = \frac{36}{36} = 1$.The slope of the tangent at $(0, 1)$ is $1$. Therefore,the slope of the normal is $-\frac{1}{1} = -1$.The equation of the normal at $(0, 1)$ is $y - 1 = -1(x - 0)$,which simplifies to $y - 1 = -x$,or $x + y = 1$.Checking the options,the point $(\frac{1}{2}, \frac{1}{2})$ satisfies the equation $x + y = 1$ because $\frac{1}{2} + \frac{1}{2} = 1$.

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:

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