If a tangent to the curve y = 6x - x^2 is par | vedclass

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(A) Given the curve equation is $y = 6x - x^2$ .....$(i)$First,we find the slope of the tangent by differentiating the curve with respect to $x$:$\fra

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$(2, 8)$

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(A) Given the curve equation is $y = 6x - x^2$ .....$(i)$First,we find the slope of the tangent by differentiating the curve with respect to $x$:$\frac{dy}{dx} = 6 - 2x$The given line is $4x - 2y - 1 = 0$,which can be rewritten as $2y = 4x - 1$,or $y = 2x - 0.5$.The slope of this line is $m = 2$.Since the tangent is parallel to the line,their slopes must be equal:$\frac{dy}{dx} = 2$$6 - 2x = 2$$2x = 4$$x = 2$Now,substitute $x = 2$ into the curve equation $(i)$ to find the $y$-coordinate:$y = 6(2) - (2)^2 = 12 - 4 = 8$Therefore,the point of tangency is $(2, 8)$.

If a tangent to the curve $y = 6x - x^2$ is parallel to the line $4x - 2y - 1 = 0$,then the point of tangency on the curve is

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