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The slope of the normal to the curve $y=3x^{2}$ at the point whose $x$ coordinate is $2$ is

\[\begin{array}{1 1}(1)\frac{1}{13}&(2)\frac{1}{14}\\(3)\frac{-1}{12}&(4)\frac{1}{12}\end{array}\]

1 Answer

$y=3x^2$
$\large\frac{dy}{dx}$$=6x$
Slope of the normal $= -\large\frac{1}{6x}$
Slope of the normal at $x=2 => \large\frac{-1}{12}$
Hence 3 is the correct answer.
answered May 19, 2014 by meena.p
 
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