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anonymous

  • one year ago

HELP ME PLEASE Which of the following exponential functions goes through the points (1, 6) and (2, 12)? f(x) = 3(2)x f(x) = 2(3)x f(x) = 3(2)−x f(x) = 2(3)−x

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  1. ybarrap
    • one year ago
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    Plug in x=1 and see if it equals 6 Plug in x=2 and see if it equals 12 Make sense?

  2. anonymous
    • one year ago
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    not really? can you guide me through it? (xD I dont want to be an answer-hogger/wanter.. I dont want the answer, just explanations :) ) @freckles

  3. anonymous
    • one year ago
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    Do you understand that coordinates are in the form (x,y)?

  4. anonymous
    • one year ago
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    yes. @BMF96

  5. anonymous
    • one year ago
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    f(x) = y

  6. anonymous
    • one year ago
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    What don't you understand?

  7. jim_thompson5910
    • one year ago
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    If \[\Large f(x) = 7(3)^x\] (for example), then what is the value of f(x) when x = 2? In other words, what is f(2) equal to?

  8. anonymous
    • one year ago
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    f(2)=441?? #CONFUSED

  9. jim_thompson5910
    • one year ago
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    Replace each x with 2 \[\Large f(x) = 7(3)^x\] \[\Large f(2) = 7(3)^2\] \[\Large f(2) = 7(9)\] \[\Large f(2) = 63\] Do you see how I got f(2) to be equal to 63?

  10. jim_thompson5910
    • one year ago
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    btw you square first and then you multiply

  11. anonymous
    • one year ago
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    ok, I forgot that rule :) (like DUHR, Bella, get a grip)

  12. jim_thompson5910
    • one year ago
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    Since \(\Large f({\color{red}{2}}) = {\color{blue}{63}}\) from my example, this means the point \(\Large ({\color{red}{x}},{\color{blue}{y}})=({\color{red}{2}},{\color{blue}{63}})\) lies on the function curve of f(x)

  13. anonymous
    • one year ago
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    okayyy....

  14. jdoe0001
    • one year ago
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    \(f(x)=y= \begin{cases} 3(2)^x\\ 2(3)^x\\ 3(2)^{-x}\\ 2(3)^{-x} \end{cases}\qquad \qquad \begin{array}{llll} x&y \\\hline\\ {\color{brown}{ 1}}&3(2)^{\color{brown}{ 1}}\\ &2(3)^{\color{brown}{ 1}}\\ &3(2)^{-{\color{brown}{ 1}}}\\ &2(3)^{-{\color{brown}{ 1}}}\\ {\color{brown}{ 2}}&3(2)^{\color{brown}{ 2}}\\ &2(3)^{\color{brown}{ 2}}\\ &3(2)^{-{\color{brown}{ 2}}}\\ &2(3)^{-{\color{brown}{ 2}}} \end{array}\)

  15. jim_thompson5910
    • one year ago
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    so what ybarrap said at the top, you plug in each x coordinate into each function and see if you get the correct corresponding y coordinates Let's say we pick choice B at random \[\Large f(x) = 2(3)^x\] The first point is (1,6). To test if this point lies on the function f(x) curve, we plug in x = 1 and see if y = 6 pops out \[\Large f(x) = 2(3)^x\] \[\Large f(1) = 2(3)^1\] \[\Large f(1) = 2(3)\] \[\Large f(1) = 6\] It does, so (1,6) is definitely on this curve. How about (2,12)? Let's check \[\Large f(x) = 2(3)^x\] \[\Large f(2) = 2(3)^2\] \[\Large f(2) = 2(9)\] \[\Large f(2) = 18\] Nope. The point (2,18) actually lies on this function curve and NOT (2,12). So we can rule out choice B.

  16. jim_thompson5910
    • one year ago
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    So what just happened was that I've proven that the function \(\Large f(x) = 2(3)^x\) does NOT go through both points (1,6) and (2,12).

  17. anonymous
    • one year ago
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    okay, so its A, C, or D lol.... so lets try to rule out A...

  18. anonymous
    • one year ago
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    @jim_thompson5910

  19. jim_thompson5910
    • one year ago
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    what did you get so far in checking choice A?

  20. anonymous
    • one year ago
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    that A is, in fact NOT the answer!?

  21. jim_thompson5910
    • one year ago
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    if x = 1, then what is f(1) ?

  22. anonymous
    • one year ago
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    f

  23. anonymous
    • one year ago
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    Wait, so it IS A!!!!!

  24. jim_thompson5910
    • one year ago
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    \[\Large f(x) = 3(2)^x\] \[\Large f(1) = 3(2)^1\] \[\Large f(1) = \underline{ \ \ \ \ \ \ \ } \text{ (fill in the blank)}\]

  25. anonymous
    • one year ago
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    6

  26. jim_thompson5910
    • one year ago
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    and how about f(2)

  27. anonymous
    • one year ago
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    12

  28. jim_thompson5910
    • one year ago
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    Good. Choice A is definitely the answer. As practice, why not go through C and D and eliminate them. With choice C, if x = 1, then what is f(x) equal to?

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