for the curve y=2(3^x)+1, determine:

Question

the horizontal asymptote
the y-intercept
range
end behaviour
also, how would i sketch this?

Answer 1

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.