Find the equations of all lines having slope | vedclass

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(D) Given the curve $y = \frac{1}{x^2 - 2x + 3}$.To find the slope of the tangent,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{-1}{

Vedclass · Accepted Answer

$2y - 1 = 0$

Vedclass · Answer

(D) Given the curve $y = \frac{1}{x^2 - 2x + 3}$.To find the slope of the tangent,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{-1}{(x^2 - 2x + 3)^2} \cdot (2x - 2) = \frac{-(2x - 2)}{(x^2 - 2x + 3)^2}$.Since the slope of the tangent is $0$,we set $\frac{dy}{dx} = 0$:$\frac{-(2x - 2)}{(x^2 - 2x + 3)^2} = 0 \implies 2x - 2 = 0 \implies x = 1$.Now,substitute $x = 1$ into the original equation to find the $y$-coordinate:$y = \frac{1}{(1)^2 - 2(1) + 3} = \frac{1}{1 - 2 + 3} = \frac{1}{2}$.The point of tangency is $(1, 1/2)$.The equation of the tangent line with slope $m = 0$ passing through $(1, 1/2)$ is:$y - \frac{1}{2} = 0(x - 1) \implies y = \frac{1}{2} \implies 2y - 1 = 0$.

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

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