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Math Help - Exponential Function form question?

  1. #1
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    Exponential Function form question?

    For this question:Exponential Function form question?-capture1.jpg My friend and I in the same class are arguing, the answer can be 1,2 all of the options or none of them. I believe it's options I and II, but my friend thinks it's II and III. Which of us is right? Or maybe it's neither of us. (Also, this is for an online practice thing that quizzes us but only gives participation points, so it's not worth anything, but it doesn't show right and wrong answers after so I won't even know what I got wrong) Thanks!
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  2. #2
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    Re: Exponential Function form question?

    What is your reasoning? You should have a reason for saying that. Suppose for I, $y = Ab^x$. Then look at the values when $x = -1$ and when $x = 1$. You have $Ab^{-1} = 20$ and $Ab = 5$. From the first equation, multiply both sides by $b$ and you find $A=20b$. Plug that in to the second equation: $(20b)b = 5$. Solving for $b$, you find $b = \pm \dfrac{1}{2}$. Obviously, that is less than 1, so it does not satisfy $b>1$.

    For II, at $x=0, y = 2$. At $x=1, y \approx 4$, and at $x=2, y\approx 8$. So, let's do the same thing. Assume $y=Ab^x$. Plugging in when $x=0,y=2$ and when $x=1,y=4$, you get: $Ab^0 = 2$, $Ab = 4$. This means $A=2$ and $2b=4$, so $b=2>1$. This does, indeed, meet the requirements of the problem.

    For III, once more, you assume $y = Ab^x$. Again, we try plugging in when $x=\pm 1$. We find two equations: $-5 = Ab^{-1}$ and $-20 = Ab$. So, from the first equation, we find $A = -5b$. Plugging that into the second, we find $-20 = (-5b)b = -5b^2$. Solving for $b$, we find $b = \pm 2$. If $b=-2$, then it is less than one, so assume $b=2$. Then, $A = -5(2) = -10$ and the equation becomes $y = -10(2)^x$. This is the second possibly correct function. Let's try the remaining points. At $x=-2$, you have $y = -10(2)^{-2} = -2.5$, which matches the chart. At $x=2$, you have $y = -10(2)^2 = -40$, and again, this matches the chart, so you might, indeed, have a function of that form.
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