posted by Rebecca .
for the curve y=2(3^x)+1, determine:
the horizontal asymptote
also, how would i sketch this?
y=3^x has a horizontal asymptote at y=0
so does y=2(3^x)
so, y=2(3^x)+1 has a h.a. at y=1.
what is y when x=0? y=2*1+1 = 3
range is all reals > 1
all exponentials look basically the same. Just graph y=a^x passes through (0,1) and (a,1)
Stretch it by 2 and shift it up 1.