The point on the curve y = 2 + sqrt{4x + 1} a | vedclass

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(C) Given the curve equation is $y = 2 + \sqrt{4x + 1}$.To find the slope of the tangent,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \fr

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$(6, 7)$

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(C) Given the curve equation is $y = 2 + \sqrt{4x + 1}$.To find the slope of the tangent,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(2 + \sqrt{4x + 1}) = 0 + \frac{1}{2\sqrt{4x + 1}} 	imes 4 = \frac{2}{\sqrt{4x + 1}}$.We are given that the slope of the tangent is $\frac{2}{5}$.So,$\frac{2}{\sqrt{4x + 1}} = \frac{2}{5}$.This implies $\sqrt{4x + 1} = 5$.Squaring both sides,we get $4x + 1 = 25$.$4x = 24$,which gives $x = 6$.Now,substitute $x = 6$ into the original equation to find $y$:$y = 2 + \sqrt{4(6) + 1} = 2 + \sqrt{25} = 2 + 5 = 7$.Thus,the required point is $(6, 7)$.

The point on the curve $y = 2 + \sqrt{4x + 1}$ at which the slope of the tangent is $\frac{2}{5}$ is:

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