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

Equation of the curve is
$y= \frac{\left ( x-7 \right )}{\left ( x-2 \right )\left ( x-3 \right )}$
put y=0 is the above equation, we get x=7
$\frac{dy}{dx}= \frac{\left ( x-2 \right )\left ( x-3 \right )-\left ( x-7 \right )\left ( 2x-5 \right )}{\left ( x-2 \right )^{2}\left ( x-3 \right )^{2}}$
the slope of the tangent at point (7,0) is
$m_{1}= \frac{dy}{dx}|\left ( 7,0 \right )= \frac{20}{400}= \frac{1}{20}$
So, the equation of the tangent is
$\frac{y-0}{x-7}= \frac{1}{20}\: \: or\: \: x-20y-7= 0$
Also the equation of the normal is
$\frac{y-0}{x-7}= -20\: \: or\: \: 20x+y-140= 0$

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