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anonymous

  • one year ago

Someone please help me with an Algebra 2 problem?

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  1. anonymous
    • one year ago
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    Your friend hands you a graph of the engine power statistics of a race car. He says, “I know this graph is f(x) = –3(x + 4)6 – 8 but I can’t remember how it is related to the graph of x6.” Explain to your friend how the graph f(x) is a translation of the graph x6.

  2. anonymous
    • one year ago
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    Start with the (x+4) part. That means the graph is translated 4 places to the left. Then the -8 part means that it start -8 blocks down, when you set x = -4, the y value is -8. The 3 means that it's getting larger faster.

  3. anonymous
    • one year ago
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    That is to the power of 6 by the way, I meant to put the little carrot in so it's: f(x) = –3(x + 4)^6 – 8

  4. anonymous
    • one year ago
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    @carina98 yes, my answers still stays the same. Except for the -3 part. It get's flipped upside down. Didn't see the negative the first time i read it.

  5. anonymous
    • one year ago
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    Okay so we are just explaining how the graph would appear with the given function? I don't necessarily understand what it means by " how it is related to the graph of x^6"

  6. anonymous
    • one year ago
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    Overall, start by shifting 4 units left, and 8 units down. Then you flip it upside down because it's negative. Then the graph gets larger quicker because of the 3.

  7. anonymous
    • one year ago
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    Don't have to worry about the x^6 part because it's simply asking what happens to the graph.

  8. anonymous
    • one year ago
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    Okay I have it graphed and I am looking at it. So I would just be explaining the change that I notice for my answer?

  9. anonymous
    • one year ago
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    Yes. You should also study a bit what the numbers do to the function. That way when another problem like this emerges, you'll be able to solve that one too.

  10. anonymous
    • one year ago
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    Okay awesome so the graph will be pointing downwards?

  11. anonymous
    • one year ago
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    Yes. The negative sign does that part.

  12. anonymous
    • one year ago
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    Okay thank you so much for walking me through this I was a bit stuck.

  13. anonymous
    • one year ago
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    No problem. Glad I could help.

  14. anonymous
    • one year ago
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    Thank you again! ♥

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