Let C be a curve given by y(x) = 1 + sqrt{4x | vedclass

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(A) Given the curve $y = 1 + \sqrt{4x - 3}$.First,find the slope of the tangent by differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{2\sqr

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$(1, 7)$

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(A) Given the curve $y = 1 + \sqrt{4x - 3}$.First,find the slope of the tangent by differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{2\sqrt{4x - 3}} 	imes 4 = \frac{2}{\sqrt{4x - 3}}$.Given the slope of the tangent is $\frac{2}{3}$,we set:$\frac{2}{\sqrt{4x - 3}} = \frac{2}{3} \Rightarrow \sqrt{4x - 3} = 3$.Squaring both sides,$4x - 3 = 9 \Rightarrow 4x = 12 \Rightarrow x = 3$.Substituting $x = 3$ into the curve equation,$y = 1 + \sqrt{4(3) - 3} = 1 + \sqrt{9} = 4$.So,the point $P$ is $(3, 4)$.The slope of the normal at $P$ is the negative reciprocal of the tangent slope:$m_{normal} = -\frac{1}{2/3} = -\frac{3}{2}$.The equation of the normal at $(3, 4)$ is:$y - 4 = -\frac{3}{2}(x - 3) \Rightarrow 2y - 8 = -3x + 9 \Rightarrow 3x + 2y - 17 = 0$.Checking the options,for $(1, 7)$: $3(1) + 2(7) - 17 = 3 + 14 - 17 = 0$. Thus,the normal passes through $(1, 7)$.

Let $C$ be a curve given by $y(x) = 1 + \sqrt{4x - 3}$,$x > \frac{3}{4}$. If $P$ is a point on $C$ such that the tangent at $P$ has slope $\frac{2}{3}$,then a point through which the normal at $P$ passes is:

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