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(B) The slope of the tangent to the curve $y = \sqrt{4x - 3} - 1$ is given by the derivative $\frac{dy}{dx}$.$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{4x -

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$(3, 2)$

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(B) The slope of the tangent to the curve $y = \sqrt{4x - 3} - 1$ is given by the derivative $\frac{dy}{dx}$.$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{4x - 3} - 1) = \frac{1}{2\sqrt{4x - 3}} 	imes 4 = \frac{2}{\sqrt{4x - 3}}$.Given that the slope is $\frac{2}{3}$,we set the derivative equal to $\frac{2}{3}$:$\frac{2}{\sqrt{4x - 3}} = \frac{2}{3}$.$\sqrt{4x - 3} = 3$.Squaring both sides,we get $4x - 3 = 9$.$4x = 12$,which implies $x = 3$.Now,substitute $x = 3$ into the original equation to find $y$:$y = \sqrt{4(3) - 3} - 1 = \sqrt{12 - 3} - 1 = \sqrt{9} - 1 = 3 - 1 = 2$.Thus,the required point is $(3, 2)$.

Find the point at which the tangent to the curve $y = \sqrt{4x - 3} - 1$ has its slope $\frac{2}{3}$.

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