Let C be a curve given by y(x)=1+sqrt{4x-3},x | vedclass

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

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$(1,7)$

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(A) Given the curve $y(x) = 1 + \sqrt{4x-3}$.First,we find the derivative to determine the slope of the tangent:$\frac{dy}{dx} = \frac{1}{2\sqrt{4x-3}} 	imes 4 = \frac{2}{\sqrt{4x-3}}$.Given the slope of the tangent at point $P$ is $\frac{2}{3}$,we set $\frac{2}{\sqrt{4x-3}} = \frac{2}{3}$.This implies $\sqrt{4x-3} = 3$,so $4x-3 = 9$,which gives $4x = 12$,or $x = 3$.Substituting $x = 3$ into the curve equation,we get $y = 1 + \sqrt{4(3)-3} = 1 + \sqrt{9} = 1 + 3 = 4$.Thus,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 $y - 4 = -\frac{3}{2}(x - 3)$.Multiplying by $2$,we get $2y - 8 = -3x + 9$,which simplifies to $3x + 2y - 17 = 0$.Checking the options:For $(1, 7)$: $3(1) + 2(7) - 17 = 3 + 14 - 17 = 0$. This satisfies the equation.Therefore,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 a slope of $\frac{2}{3}$,then a point through which the normal at $P$ passes is:

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