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

Explanation

The equation of the given curve is `y=1/(x^2-2x+3`

The slope of the tangent to the given curve at any point (x,y) is given by,

`dy/dx=-(2x-2)/((x^2-2x+3)^3)=-(2(x-1))/((x^2-2x+3)^3)`

If the slope of the tangent is 0, then we have:

`-(2(x-1))/((x^2-2x+3)^3)=0`

When `x=1` , `y=1/2`

the equation of the tangent through `(1,1/2)` is given by

`y-1/2=0(x-1)impliesy=1/2`