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

$(A) \; (-3, -2)$
$(B)\; (3, -2)$
$(C)\; (3, 2)$
$(D)\; (-3, 2)$

Given $y = \sqrt {4x-3} -1$
$\frac{dy}{dx} = \frac{1}{2} (4x-3)^{\frac{-1}{2}} . 4 = \frac{2}{\sqrt{4x-3}}$
The slpe is given to be $\frac{2}{3}$
\begin{align*}\therefore \frac{2}{\sqrt{4x-3}} =\frac{2}{3} \implies 4x - 3 & = 9 \\ \therefore x & = 3\end{align*}
$y = \sqrt{4x-3} -1.$ Hence where $x = 3, y = \sqrt{4(3) - 3} -1 = 2$