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anonymous

  • one year ago

Pre-Calculus exponential functions help?

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    make an x/y table to get points and plot

  3. anonymous
    • one year ago
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    |dw:1443201477463:dw|

  4. anonymous
    • one year ago
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    where did the 1/4 and 1/2 come from?

  5. anonymous
    • one year ago
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    \[2^{-1} = \frac{ 1 }{ 2^1 }=\frac{ 1 }{ 2 }\]\[2^{-2}=\frac{ 1 }{ 2^2 }=\frac{ 1 }{ 4 }\]

  6. anonymous
    • one year ago
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    Oh okay

  7. anonymous
    • one year ago
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    So what do I do with the points from the table then? Plot them?

  8. anonymous
    • one year ago
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    yes

  9. anonymous
    • one year ago
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    But that table was only for 2^x, right? I still need a table for 2^-x?

  10. anonymous
    • one year ago
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    yes

  11. anonymous
    • one year ago
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    do you know your transformations?

  12. anonymous
    • one year ago
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    kind of

  13. anonymous
    • one year ago
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    so what if g(x) = f(-x) what is g compared to f? in other words, g is what transformation of f?

  14. anonymous
    • one year ago
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    reflection?

  15. anonymous
    • one year ago
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    about what?

  16. anonymous
    • one year ago
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    the y axis

  17. anonymous
    • one year ago
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    very good! so f(x) = 2^x and g(x) = 2^(-x) hint, hint, wink, wink... if you know one and you understand transformations, you really know both

  18. anonymous
    • one year ago
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    so after I graph 2^x, then I just reflect it about the y axis for the graph of 2^-x?

  19. anonymous
    • one year ago
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    right?

  20. anonymous
    • one year ago
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    So they're exponential functions because they increase or decrease exponentially?

  21. anonymous
    • one year ago
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    yes but where is the variable x?

  22. anonymous
    • one year ago
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    I don't understand what you mean

  23. anonymous
    • one year ago
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    \(f\left(x\right) = y = 2^x\)

  24. anonymous
    • one year ago
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    x is in the exponent

  25. anonymous
    • one year ago
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    and that's what is changing

  26. anonymous
    • one year ago
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    thus an "exponential" function

  27. anonymous
    • one year ago
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    ohhh okay!

  28. anonymous
    • one year ago
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    so what if we're given the graph and need to find the function? (I have an example)

  29. anonymous
    • one year ago
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    http://www.purplemath.com/modules/expofcns.htm exponential functions are also characterized by multiplying in a recursive relation... \[f(x)=2^x=2\cdot 2^{x-1}=2\cdot f(x-1)\] so if you have a graph, divide f(x)/f(x-1) to find the base

  30. anonymous
    • one year ago
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    how do we know what f(x) is?

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  31. anonymous
    • one year ago
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    f(2) = 9, what is f(1)?

  32. anonymous
    • one year ago
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    2?

  33. anonymous
    • one year ago
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    that's what it looks like to me... so f(2)/f(1) = base

  34. anonymous
    • one year ago
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    so 9/2 is the base?

  35. anonymous
    • one year ago
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    that's what i think

  36. anonymous
    • one year ago
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    okay, then what's the exponent part?

  37. anonymous
    • one year ago
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    whoa... wait

  38. anonymous
    • one year ago
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    what?

  39. anonymous
    • one year ago
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    f(1) = 2 <- this one is not correct f(2) = 9 f(0) = 1 it's 2 tick marks but how much is each tick mark?

  40. anonymous
    • one year ago
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    I thought it showed that each mark is 1...

  41. anonymous
    • one year ago
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    if you ask me, f(1) = 3 look closely at the spacing of the vertical tick marks. the first tick is about half of the subsequent ticks

  42. anonymous
    • one year ago
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    there are 5 marks to 9... 1 + 2 + 2 + 2 + 2 = 9

  43. anonymous
    • one year ago
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    besides (9/2)^1 = 9/2 \(\ne\) 2 and (9/2)^2 = 81/4 \(\ne\) 9

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