**Question:**

Find the equation of all lines of slope zero and that is tangent to the curve $y=\frac{1}{x^{2}-2 x+3}$.

**Solution:**

finding the slope of the tangent by differentiating the curve

$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{(2 \mathrm{x}-2)}{\left(\mathrm{x}^{2}-2 \mathrm{x}+3\right)}$

Now according to question, the slope of all tangents is equal to 0 , so

$-\frac{(2 x-2)}{\left(x^{2}-2 x+3\right)}=0$

Therefore the only possible solution is $x=1$

since this point lies on the curve, we can find $y$ by substituting $x$

$y=\frac{1}{1-2+3}$

$y=\frac{1}{2}$

equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$

$y-\frac{1}{2}=0(x-1)$

$y=\frac{1}{2}$