Question 682699
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^2\ -\ x\ +\ 1\ -\ \ln(x\ +\ 4)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y'\ =\ \frac{dy}{dx}\ =\ 2x\ -\ 1\ -\ \frac{1}{x\ +\ 4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y'(-3)\ =\ -6\ -\ 1\ -\ 1\ =\ -8]


Use the point-slope form


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 13\ =\ -8(x\ +\ 3)]


Simplify as required.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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