The point at which the tangent to the curve y=√(4x-3)-1 has its slope…

The correct answer is Option (1) → (3, 2)

$y=\sqrt{4x-3}-1$  ...(1)

so $(y+1)^2=4x-3$ ....(2)

differentiating wrt x

$2(y+1)\frac{dy}{dx}=4$ so $\frac{dy}{dx}=\frac{2}{3}$

so $2(y+1)\frac{2}{3}=4⇒y=2$

From (2) → $9=4x-3$

$⇒x=3$

Point (3, 2)