The tangent to the curve y = frac{1}{x^2 + 2x | vedclass

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(A) Given the curve $y = \frac{1}{x^2 + 2x + 5}$.To find the tangent parallel to the $X$-axis,we set the derivative $\frac{dy}{dx} = 0$.Using the chai

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$y = 1/4$

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(A) Given the curve $y = \frac{1}{x^2 + 2x + 5}$.To find the tangent parallel to the $X$-axis,we set the derivative $\frac{dy}{dx} = 0$.Using the chain rule,$\frac{dy}{dx} = -\frac{1}{(x^2 + 2x + 5)^2} \cdot (2x + 2)$.Setting $\frac{dy}{dx} = 0$ gives $-(2x + 2) = 0$,which implies $x = -1$.Now,substitute $x = -1$ into the original equation to find the $y$-coordinate:$y = \frac{1}{(-1)^2 + 2(-1) + 5} = \frac{1}{1 - 2 + 5} = \frac{1}{4}$.Since the tangent is parallel to the $X$-axis,its equation is $y = 	ext{constant}$,which is $y = 1/4$.

The tangent to the curve $y = \frac{1}{x^2 + 2x + 5}$ which is parallel to the $X$-axis is:

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