Find the equation of the line having slope 0 | vedclass

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Question

(A) The equation of the given curve is $y=\frac{1}{x^{2}-2x+3}$.The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of the given curve is $y=\frac{1}{x^{2}-2x+3}$.The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.Using the chain rule,$\frac{dy}{dx} = -\frac{1}{(x^{2}-2x+3)^{2}} \cdot (2x-2) = \frac{-2(x-1)}{(x^{2}-2x+3)^{2}}$.Given that the slope of the tangent is $0$,we set $\frac{dy}{dx} = 0$:$\frac{-2(x-1)}{(x^{2}-2x+3)^{2}} = 0$.This implies $-2(x-1) = 0$,so $x = 1$.Now,find the corresponding $y$-coordinate by substituting $x=1$ into the curve equation:$y = \frac{1}{(1)^{2}-2(1)+3} = \frac{1}{1-2+3} = \frac{1}{2}$.The point of tangency is $(1, \frac{1}{2})$.The equation of a line with slope $m=0$ passing through $(x_0, y_0)$ is $y - y_0 = m(x - x_0)$.Substituting the values,$y - \frac{1}{2} = 0(x - 1)$,which simplifies to $y = \frac{1}{2}$.

Find the equation of the line having slope $0$ which is tangent to the curve $y=\frac{1}{x^{2}-2x+3}$.

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