Question 1053129
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Find the value of the first derivative of the function at *[tex \Large x\ =\ -2]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \left.\frac{dy}{dx}\right|_{\small{x\,=\,-2}}\LARGE\ =\ -2(-2)\ +\ 4 =\ 8]


And the value of the function at -2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(-2)\ =\ -(-2)^2\ +\ 4(-2)\ =\ -12]


So the normal to the curve at -2 has a slope of *[tex \Large -\frac{1}{8}] and passes through the point *[tex \Large (-2,-12)]


Using the Point-Slope form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ 12\ =\ -\frac{1}{8}\left(x\ +\ 2)].


You can put it in whatever form you like or whatever form your instructor requires.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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