The abscissae of the points of the curve y = | vedclass

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Question

(A) Given the curve equation: $y = x(x - 2)(x - 4)$.Expanding the expression: $y = x(x^2 - 6x + 8) = x^3 - 6x^2 + 8x$.To find the slope of the tangent

Vedclass · Accepted Answer

$x = 2 \pm \frac{2}{\sqrt{3}}$

Vedclass · Answer

(A) Given the curve equation: $y = x(x - 2)(x - 4)$.Expanding the expression: $y = x(x^2 - 6x + 8) = x^3 - 6x^2 + 8x$.To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 3x^2 - 12x + 8$.For the tangent to be parallel to the $x$-axis,the slope must be zero: $\frac{dy}{dx} = 0$.Setting the derivative to zero: $3x^2 - 12x + 8 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(8)}}{2(3)} = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm \sqrt{48}}{6}$.Simplifying the expression: $x = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2\sqrt{3}}{3} = 2 \pm \frac{2}{\sqrt{3}}$.

The abscissae of the points of the curve $y = x(x - 2)(x - 4)$ where the tangents are parallel to the $x$-axis are obtained as:

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