Consider the curve given by the equation y^(3)-8y^(2)+16 y-x^(2)+2x=0". " It can be shown that (dy)/(dx)=(2(x-1))/((3y-4)(y-4))". " Find the y-coordinate of the point where the line tangent to the curve is the y-axis. y=

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Consider the curve given by the equation  y^(3)-8y^(2)+16 y-x^(2)+2x=0". " It can be shown that  (dy)/(dx)=(2(x-1))/((3y-4)(y-4))". " Find the  y-coordinate of the point where the line tangent to the curve is the  y-axis.  y=

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Find Tangent Point: To find the y-coordinate of the point where the line tangent to the curve is the y-axis, we need to find the point where the derivative (dy/dx) is undefined or zero, since the slope of the tangent line to the y-axis is vertical (undefined slope).Derivative Calculation: The derivative of y with respect to x is given by (dy/dx) = (2(x-1))/((3y-4)(y-4)). For the tangent to be the y-axis, the numerator of the derivative must be zero, because a zero numerator over a non-zero denominator gives a slope of zero, which corresponds to a horizontal line, not a vertical one. Therefore, we need to find where the denominator is zero, which would make the slope undefined.Solve for Undefined Slope: Set the denominator of the derivative equal to zero to find the y-coordinate where the slope is undefined: (3y - 4)(y - 4) = 0.Check Y-Values: Solve the equation (3y - 4)(y - 4) = 0 by setting each factor equal to zero: 3y - 4 = 0 or y - 4 = 0.Substitute x=0: Solving 3y - 4 = 0 gives y = 4/3, and solving y - 4 = 0 gives y = 4.Factor Out Equation: We need to check which of these y-values, if any, correspond to the point where the tangent is the y-axis. We know that at the y-axis, x = 0. So we substitute x = 0 into the original equation to see if either of these y-values satisfy the equation.Final Y-Coordinate: Substitute x = 0 into the original equation y^3 - 8y^2 + 16y - x^2 + 2x = 0 to get y^3 - 8y^2 + 16y = 0.Factor out y from the equation y^3 - 8y^2 + 16y = 0 to get y(y^2 - 8y + 16) = 0.The factored form of y^2 - 8y + 16 is (y - 4)^2, so the equation becomes y(y - 4)^2 = 0.Set each factor equal to zero: y = 0 or (y - 4)^2 = 0. Solving (y - 4)^2 = 0 gives y = 4.The y-value that satisfies the original equation when x = 0 is y = 4. This is the y-coordinate of the point where the line tangent to the curve is the y-axis.

Consider the curve given by the equation $y^{3}-8 y^{2}+16 y-x^{2}+2 x=0 \text {. }$ It can be shown that $\frac{d y}{d x}=\frac{2(x-1)}{(3 y-4)(y-4)} \text {. }$ $\newline$ Find the $y$ -coordinate of the point where the line tangent to the curve is the $y$ -axis. $\newline$ $y=$

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