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.

 

 

 

 
 
 
 
 

Answers (1)

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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