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Find the equation of the normal line and the tangent line to the indicated curve at the given point. $$y=3x^2 \text{ at } (-4,48)$$

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closed as off-topic by Shailesh, Davide Giraudo, Claude Leibovici, Namaste, Juniven Apr 24 '17 at 16:10

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We have $y'=6x$. At $x=-4$, $y'=-24$. So the slope of the normal line is $\frac{1}{24}$. The normal line is given by $$y-48=\frac{1}{24}(x+4)$$ that is, $$24y-48(24)=x+4$$ The tangent line is given by $$y-48=-24(x+4)$$

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