Find the point on the curve y = sqrt{4x - 3} | vedclass

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(C) The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.Given $y = \sqrt{4x - 3} - 1$,we differenti

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$(3, 2)$

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(C) The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.Given $y = \sqrt{4x - 3} - 1$,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{2\sqrt{4x - 3}} 	imes 4 = \frac{2}{\sqrt{4x - 3}}$.We are given that the slope of the tangent is $\frac{2}{3}$.Equating the derivative to the given slope:$\frac{2}{\sqrt{4x - 3}} = \frac{2}{3}$$\sqrt{4x - 3} = 3$Squaring both sides:$4x - 3 = 9$$4x = 12$$x = 3$.Now,substitute $x = 3$ into the original equation of the curve to find $y$:$y = \sqrt{4(3) - 3} - 1$$y = \sqrt{12 - 3} - 1$$y = \sqrt{9} - 1$$y = 3 - 1 = 2$.Thus,the required point is $(3, 2)$.

Find the point on the curve $y = \sqrt{4x - 3} - 1$ where the tangent has a slope of $\frac{2}{3}$.

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