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Math Help - hard time with graph problems

  1. #1
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    hard time with graph problems

    Identify the exponential function of the form
    f (x) = a(2 x ) + b whose
    graph is shown in the figure.
    a. f (x) = 3(2 x )
    b.


    f (x) = 2 x 3
    c.


    f (x) = 2(2 x ) 4
    d. f (x) = 2 x 2

    The graph is attached a a work file.

    thanks for any help

    Attached Files Attached Files
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by wvmcanelly@cableone.net View Post
    Identify the exponential function of the form
    f (x) = a(2 x ) + b whose
    graph is shown in the figure.
    a. f (x) = 3(2 x )
    b.


    f (x) = 2 x 3
    c.


    f (x) = 2(2 x ) 4
    d. f (x) = 2 x 2

    The graph is attached a a work file.

    thanks for any help

    what you have in brackets, are those suppsoed to be powers? when you say 3(2 x) do you mean 3 \cdot 2^x \mbox { or } 3^{2x} ?
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  3. #3
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    My apologies,

    I'm half asleep.

    the answer options should be

    f(x) = 3(2^x)
    f(x) = 2^x-3
    f(x) = 2(2^x) - 4
    f(x) = 2^x - 2
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by wvmcanelly@cableone.net View Post
    My apologies,

    I'm half asleep.

    the answer options should be

    f(x) = 3(2^x)
    f(x) = 2^x-3
    f(x) = 2(2^x) - 4
    f(x) = 2^x - 2
    now we could go into all the theories about how to find asymptotes and how an exponential function behaves and whatnot, but here, i think it is simplest to just check the points given

    on the graph, we have the points (0, -2) and (2, 1). these points will work for only one of the choices.


    (0, - 2) means, when x is 0, y is -2, so let's test that

    plug in x = 0 in the first, we get y = 3, so thats not our guy, so (a) is out!

    plug in x = 0 in the second, we get y = -2, so this MAY be our guy, (b) is still in

    plug in x = 0 in the third, we get y = -2, so (c) is still in

    plug in x = 0 in the fourth, we get y = -1, so (d) is out!



    Now just check (b) and (c) for the second point.

    (2,1) means, when x = 2, y = 1

    so for (b), plug in x = 2, we get y = 1, ah, this seems to be it!

    so for (c), plug in x = 2, we get y = 4, this is not our guy!

    so the answer is (c), y = 2^x - 3



    NOTE: we could also realize that 2^x has a horizontal asymptote at y = 0. adding -3 shifts eveything down 3 units, so the asymptote becomes y = -3. no other graph is like that, so we could have gotten the answer that way
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  5. #5
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    Thanks, I really have a hard time with graphs. I was really stuck on this one and I appreciate you explaining the problem
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