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iwanttogotostanford

  • one year ago

ACT PREP Q/ PLEASE HELPPP: ACT PREP Q/ PLEASE HELP: The functions f(x) = –(x + 4)^2 + 2 and g(x) = (x − 2)^2 − 2 have been rewritten using the completing-the-square method. Is the vertex for each function a minimum or a maximum? Explain your reasoning for each function.

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  1. iwanttogotostanford
    • one year ago
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    @zepdrix

  2. zepdrix
    • one year ago
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    \(\large\rm y=x^2\) is a parabola opening `upward`.|dw:1443654118776:dw|

  3. zepdrix
    • one year ago
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    |dw:1443654152110:dw|So the vertex is a minimum. It is lower than any other point on the curve.

  4. zepdrix
    • one year ago
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    \(\large\rm y=-x^2\) is a parabola opening downward.|dw:1443654195359:dw|

  5. zepdrix
    • one year ago
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    |dw:1443654246434:dw|And we can clearly see that the vertex is the highest point on the curve, the maximum.

  6. zepdrix
    • one year ago
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    Hmmmm so what do you think? :d how does that apply to f(x) and g(x)

  7. iwanttogotostanford
    • one year ago
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    im not sure, i don't know how to do this at all- i am studying for the ACT and was so clueless on this one!

  8. iwanttogotostanford
    • one year ago
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    I have not learned this yet and this was one of the harder questions for the ACT practice

  9. iwanttogotostanford
    • one year ago
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    @zepdrix

  10. zepdrix
    • one year ago
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    well use my examples :O and try to guess at least

  11. iwanttogotostanford
    • one year ago
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    So I was hoping for a step by step so I could use it for learning! :-)

  12. zepdrix
    • one year ago
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    u -_-

  13. iwanttogotostanford
    • one year ago
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    What??! I have an ACT prep book right in front of me and I am clueless! Its not for a grade

  14. zepdrix
    • one year ago
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    no, im saying, give ME a guess. lol

  15. zepdrix
    • one year ago
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    do you understand how to find the vertex when a function is given in this form? :)

  16. iwanttogotostanford
    • one year ago
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    ok, well I think the vertex is a minimum and NOPE

  17. iwanttogotostanford
    • one year ago
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    I am learning algebra 2 right noe but I have the ACT this october

  18. zepdrix
    • one year ago
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    well unfortunately, this is two questions, not just one. It's asking about two separate functions, f(x) and g(x).

  19. iwanttogotostanford
    • one year ago
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    i know! I need helpp

  20. iwanttogotostanford
    • one year ago
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    please

  21. zepdrix
    • one year ago
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    the vertex form of a parabola looks like this:\[\large\rm y=(x-\color{#3366CF}{h})^2+\color{#3366CF}{k}\]Where the vertex is located at \(\large\rm (\color{#3366CF }{h},~\color{#3366CF }{k})\). But what is interesting is.... the location of the vertex actually has no impact on whether it is a maximum or minimum!

  22. iwanttogotostanford
    • one year ago
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    huh thats interesting!

  23. zepdrix
    • one year ago
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    So for our function:\[\large\rm f(x)=-(x-\color{#3366CF}{(-4)})^2+\color{#3366CF}{2}\]We can see that our vertex is located at \(\large\rm (\color{#3366CF }{-4},~\color{#3366CF }{2})\). But this doesn't matter so much. All we really care about, is whether or not there is a `negative sign` in front of the squared term.

  24. iwanttogotostanford
    • one year ago
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    oh ok

  25. zepdrix
    • one year ago
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    `Negative sign` means the parabola opens down, means vertex is a `maximum`. `positive sign` means the parabola opens up, meaning the vertex is a `minimum`. For f(x), the squared term has a `negative` in front. So our vertex corresponds to a `maximum`. How bout g(x)? Any ideas? :o

  26. iwanttogotostanford
    • one year ago
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    @zepdrix minimum right? for g(x)

  27. zepdrix
    • one year ago
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    \[\large\rm g(x)=(x-2)^2-2\]Mmmm ok good! There are a lot of negatives floating around in g, so it's easy to get confused, but what's important is that the square is positive,\[\large\rm g(x)=+(x-2)^2-2\]So the vertex of g(x) corresponds to a minimum. yay team \c:/

  28. zepdrix
    • one year ago
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    XD

  29. iwanttogotostanford
    • one year ago
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    @zepdrix

  30. iwanttogotostanford
    • one year ago
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    @zepdrix

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