Question

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

Answer 1

given, $\bf{y=\frac{1}{x^2-2x+3}}$
differentiate y with respect to x,
dy/dx = -1/(x² - 2x + 3) × d(x² - 2x + 3)/dx
= -1/(x² - 2x + 3) × (2x - 2)
= 2(1 - x)/(x² - 2x + 3)

A/C to question,
equations of all lines having slope 0 which are tangent to the curve y = 1/(x² - 2x + 3)
so, 2(1 - x)/(x² - 2x + 3) = 0 => x = 1

put x = 1 in y = 1/(x² - 2x + 3)
e.g., y = 1/(1² - 2.1 + 3) = 1/(1 - 2 + 3) = 1/2

equation of line passing through (1,1/2) and slope is 0 :
(y - 1/2) = 0(x - 1) => $\bf{y = 1/2}$