Question 1164742
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If f has a vertical asymptote at *[tex \Large x\ =\ 1] then *[tex \Large x\ -\ 1] must be a factor of *[tex \Large 2x^2\ +\ ax\ +\ b] and if f has a removable discontinuity at *[tex \Large x\ =\ -2] then *[tex \Large x\ +\ 2] must be a factor of both the numerator and the denominator functions.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 1)(x\ +\ 2)\ =\ x^2\ +\ x\ -\ 2]


So we need another factor of 2 for the denominator so the denominator function is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 2x\ -\ 4]


Demonstrating that *[tex \Large a\ =\ 2] and *[tex \Large b\ =\ -4]


Since *[tex \Large x\ +\ 2] is also a factor of *[tex \Large cx\ -\ 5x^2], if you factor *[tex \Large -5x] from the numerator function you get *[tex \Large -5x(-\frac{c}{5}\ +\ x)], so *[tex \Large -\frac{c}{5}\ +\ x\ =\ x\ +\ 2] means that *[tex \Large c\ =\ -10]


The lead coefficient of the numerator is -5 and the lead coefficient of the denominator is 2, therefore the equation of the horizontal asymptote is *[tex \Large y\ =\ -\frac{5}{2}]


For the last part, remove the *[tex \Large x\ +\ 2] factor from both the numerator and denominator and then evaluate at -2

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