Question 1092233: Consider Y= 2e^(-x)+1
a. Find Y when x=0, +/-1, +/-2
b. Discuss Y as x--> +/- infinity
c. Sketch the graph of y=2e^(-x)+1
d. State the equation of any asymptote
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! When x=0, the value is 2(1)+1=3; at 1 it is (2/e)+1, or about 1.74; at -1 it is 2e+1, or about 6.4,
At 2, it is 2/e^2 +1 or about 1.27; at -2, it is 2e^2+1 or about 15.8
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when x goes to +oo , y goes to +1. y=1 is the asymptote.
When x goes to -oo, y goes to +oo
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