Question : Find the eq of tangent and normal to the curve

$\displaystyle y(x-2)(x-3)-x+7 = 0 $

at the point where the curve meets the x-axis

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Solution:

$\displaystyle y = \frac{x-7}{x^2 - 5x +6 }$.............Is that correct?

$\displaystyle \therefore \frac{dy}{dx} = \frac{-x^2 + 14x -29}{(x^2 - 5x +6)^2}$...................Is that correct?????