Question

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

Answer 1

The equation of the given curve is $y=\frac{1}{x^2-2x+3}$.
The slope of the tangent to the given curve at any point $(x,y)$ is given by,
$\frac{dy}{dx}=\frac{-(2x-2)}{(x^2-2x+3{)}^3}=\frac{-2(x-1)}{(x^2-2x+3{)}^3}$
If the slope of the tangent is $0$, then we have:
$\frac{-2(x-1)}{(x^2-2x+3{)}^3}=0$
When $x=1,y=\frac{1}{1-2+3}=\frac{1}{2}$
$\therefore$ the equation of the tangent through $(1,\frac{1}{2})$ is given by,
$y-\frac{1}{2}=0(x-1)\Rightarrow y=\frac{1}{2}$