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=\(\frac12\)