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Find the equations of all lines having slope 0 which are tangent to the curve y =1 / x ^ 2 - 2 x + 3

12) Find the equations of all lines having slope 0 which are tangent to the curve
y = \frac{1}{x^2 - 2 x +3 }

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We know that the slope of the tangent  at a point on the given curve is given by  \frac{dy}{dx}

Given the equation of the curve as
y = \frac{1}{x^2 - 2x + 3}
\frac{dy}{dx} = \frac{-(2x-2)}{(x^2-2x+3)^2}
It is given thta slope is 0
So,
\frac{-(2x-2)}{(x^2 - 2x +3)^2} = 0 \Rightarrow 2x-2 = 0 = x = 1
Now, when x = 1 , y = \frac{1}{x^2-2x+3} = \frac{1}{1^2-2(1)+3} = \frac{1}{1-2+3} =\frac{1}{2}

Hence, the coordinates are \left ( 1,\frac{1}{2} \right )
Equation of line passing through \left ( 1,\frac{1}{2} \right ) and having slope = 0 is
y = mx + c
1/2 = 0 X 1 + c
c = 1/2
Now equation of line is 
y = \frac{1}{2}
 

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