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,

$\frac{dy}{dx}$ =$\frac{(-2x-2)}{(x^2-2x+3)^2}$ =$\frac{-2(x-1)}{(x^2-2x+3)^2}$

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

⇒ $\frac{-2(x-1)}{(x^2-2x+3)^2}$=0

⇒ -2(x-1)=0

⇒ x=1

When x = 1, y=$\frac{1}{1-2+3}$ =$\frac12$.

∴The equation of the tangent through(1,$\frac12$) is given by,

y-$\frac12$=0(x-1)

y-$\frac12$=0

y=$\frac12$

Hence, the equation of the required line is y=$\frac12$