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The point on the curve $y=3x^{2}-6x-4$ at which the tangent is parallel to the $x$ -axis is

\[\begin {array}{1 1}(1)\bigg(\frac{5}{2},\frac{-17}{2}\bigg)&(2)\bigg(\frac{-5}{2},\frac{-17}{2}\bigg)\\(3)\bigg(\frac{-5}{2},\frac{17}{2}\bigg)&(4)\bigg(\frac{3}{2},\frac{-17}{2}\bigg)\end{array}\]

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$y=2x^2-6x-4$
$\large\frac{dy}{dx}$$=4x-6$
Let $(x_1,y_1)$ be the point on the curve at which the tangent is parallel to x-axis
$\large\frac{dy}{dx} $$/(x_1,y_1)=4x_1-6$
Since the tangent is parallel to x-axis
$\large\frac{dy}{dx} $$/(x_1,y_1)=0$
$4x_1-6=0=> x_1=\large\frac{3}{2}$
$(x_1,y_1)$ is a point on the curve
$\therefore y_1=2x_1^2-6x_1-4$
$\qquad= 2 \times \large\frac{9}{4} $$ -6 \times \large\frac{3}{2}$$-4$
$\qquad= \large\frac{9}{2}-\frac{18}{2}$$-4$
$\qquad= \large\frac{9-18-8}{2}$
$\qquad= \large\frac{-17}{2}$
The point is $\bigg(\frac{3}{2},\frac{-17}{2}\bigg)$
Hence 4 is the correct answer
answered May 19, 2014 by meena.p
 

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