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brodybarnes35

  • one year ago

The graphs of f(x) and g(x) are shown below: graph of function f(x) open upward and has its vertex at (-7, 0). Graph of function g(x) opens upward and has its vertex at (-7,-2). If f(x) = (x + 7)^2, which of the following is g(x) based on the translation? g(x) = (x + 9)^2 g(x) = (x + 5)^2 g(x) = (x + 7)^2 + 2 g(x) = (x + 7)^2 − 2

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  1. brodybarnes35
    • one year ago
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    @Hero

  2. brodybarnes35
    • one year ago
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    some one help FAN AND MEDAL

  3. Hero
    • one year ago
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    Where are the "graphs" of f(x) and g(x)?

  4. brodybarnes35
    • one year ago
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  5. brodybarnes35
    • one year ago
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    Srry thats the graph...

  6. brodybarnes35
    • one year ago
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    @Hero

  7. Hero
    • one year ago
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    So what are your thoughts regarding this problem? How much of this do you already understand and how far have you gotten in your attempts to find the equation for g(x)?

  8. brodybarnes35
    • one year ago
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    i have attempted many times and i do not know like anything about this problem or how to do it

  9. brodybarnes35
    • one year ago
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    if you could explain that would be awesome

  10. Hero
    • one year ago
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    Well, let's start with figuring out where the points are in relation to each graph. We know that (-7,0) is on the graph of f(x). And (-7,-2) is on the graph of g(x). Do you think you can describe where (-7,-2) is in relation to (-7,0)?

  11. brodybarnes35
    • one year ago
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    the x values

  12. brodybarnes35
    • one year ago
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    ?

  13. Hero
    • one year ago
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    If you were to plot each point on an xy plane, you'd see that (-7,-2) is two units below (-7,0).

  14. brodybarnes35
    • one year ago
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    ahhhh so D?

  15. Hero
    • one year ago
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    According to the question asked in the problem, g(x) is a "translation" of f(x) so all we need to do is find the equivalent of that for the equation of f(x).

  16. Hero
    • one year ago
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    Yes, D is correct.

  17. brodybarnes35
    • one year ago
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    Thanks can u help with like a couple more?

  18. brodybarnes35
    • one year ago
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    @Hero please

  19. Hero
    • one year ago
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    Close this one and post your next question as a new one.

  20. brodybarnes35
    • one year ago
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    ok

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