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for the curve y=2(3^x)+1, determine:
the horizontal asymptote
the y-intercept
end behaviour
also, how would i sketch this?

  • Math -

    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.

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