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

Explanation

Slope of tangent = 0
Equation of curve
`y= 1/(x^(2)-2x+3)` ….(1)
` rArr (dy)/(dx) = (-1)/((x^(2)-2x+3)^(2)) d/(dx) (x^(2)-2x+3)`
`=(-2(x-1))/((x^(2)-2x+3)^(2))`
Slope of tangent at point `(x, y) = (-2(x-1))/((x^(2)-2x+3)^(2))`
`:. (-2(x-1))/((x^(2)-2x+3)^(2))=0`
` rArr x = 1`
put x = 1 in equation (1)
` y = 1/(1-2+3) = 1/2`
equation of tangent at point ` (1,1/2)` ,brgt ` y - 1/2 = 0 (x-1)`
` rArr y = 1/2`