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anonymous

  • 2 years ago

ANYONE ?? Graphing Quadratics HELPP ! (Question 1 and answer is attached) Graph one of your 2nd degree functions from question 1. Identify which function you used and the key features of your graph. Explain how to find them algebraically.

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  1. anonymous
    • 2 years ago
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  2. anonymous
    • 2 years ago
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    the answer is message me!!

  3. anonymous
    • 2 years ago
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    @amistre64 @Luigi0210 @Compassionate

  4. anonymous
    • 2 years ago
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    @AllTehMaffs @johnweldon1993 ?

  5. johnweldon1993
    • 2 years ago
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    We'll work with the f(x) f(x) = (x - 2)(x + 2) Alright it's important to remember this is the FACTORED version of a function...what this shows is where the zeros (x-intercepts) of the function are... What value(s) of 'x' make (x - 2)(x + 2) = 0 ??

  6. anonymous
    • 2 years ago
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    Oh God .. I have NO idea .... I don't get any of this , I had help on question 1 as well :/

  7. johnweldon1993
    • 2 years ago
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    Alright no problem we'll walk through it :)

  8. anonymous
    • 2 years ago
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    Oh my goodness thank you so much ..

  9. johnweldon1993
    • 2 years ago
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    Okay....so we have f(x) = (x - 2)(x + 2) Like I said above...seeing a function in this factored form...lets us find out where the zeros (also known as the x-intercepts) of the graph are... This is done by setting this form = to 0...so (x - 2)(x + 2) = 0 Okay...we know that anything times 0 = 0 right?

  10. johnweldon1993
    • 2 years ago
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    Since we know...we can see that there are 2 values of 'x' that would make this function = 0... (x - 2)(x + 2) = 0 Well...we know that if x = 2.... (2 - 2)(2 + 2) = 0 ( 0) (4) = 0 Well indeed as we know 0 times 4 DOES = 0...so x = 2 would be one x-intercept... To find the other...if x = -2 we have (-2 - 2)(-2 + 2) = 0 ( -4 ) (0) = 0 Again it is true that -4 times 0 = 0...so -2 is ANOTHER x-intercept for the graph...

  11. johnweldon1993
    • 2 years ago
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    so far we just have the 2 x-intercepts...this looks like |dw:1386883312282:dw|

  12. anonymous
    • 2 years ago
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    So is that all it's asking me to do ? :o

  13. johnweldon1993
    • 2 years ago
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    Not quite...almost done though :) are you with me through what I've done so far?

  14. anonymous
    • 2 years ago
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    Ehh , kind of ... It's just a lot of information

  15. johnweldon1993
    • 2 years ago
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    I know, trying to explain it well :/ lol just try to remember in this factored form...set the equation = to 0....and solve for 'x' (x + 2)(x - 2) = 0 We knew either x+2 had to equal 0... or we know that x-2 had to equal 0... So that's how we solved for the 2 x-intercepts... maybe a little summary better?

  16. anonymous
    • 2 years ago
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    That was a lot easier to understand :)

  17. johnweldon1993
    • 2 years ago
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    Awesome! :) Okay....so now...do you know how to expand this factored form?? to make it look something like x² + bx + c

  18. anonymous
    • 2 years ago
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    Omg no :'(

  19. johnweldon1993
    • 2 years ago
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    Don't get down...this is a pretty simple process...stay with me :)

  20. johnweldon1993
    • 2 years ago
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    We have... (x - 2)(x + 2) in order to expand this to a quadratic function...we use the FOIL method... it says... Take the first term in the first parenthesis...and multiply it by each term in the second parenthesis... Redo this with the second term in the first parenthesis... Sound understandable? I'm going to do it out anyways lol

  21. johnweldon1993
    • 2 years ago
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    so..."Take the first term in the first parenthesis...and multiply it by each term in the second parenthesis..." The first term in the first parenthesis (x - 2) is 'x' right? So we multiply it by each term in the second parenthesis (x + 2) so x times x = ? x times 2 = ? Now we just redo this for the second term in the first parenthesis (x - 2) which would be -2 so -2 times x = ? -2 times 2 = ? Fill in those question marks for me :)

  22. anonymous
    • 2 years ago
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    Wait fill out the first 2 question marks or the second 2 ..?

  23. johnweldon1993
    • 2 years ago
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    Both :)

  24. anonymous
    • 2 years ago
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    Ok well the first 2 are: x times x = 1 ? and x times 2 = 2x ? Second 2: -2 times x = -2x ? and -2 times 2 = -4 ?

  25. johnweldon1993
    • 2 years ago
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    Oh so close...everything is right but what is x times x? x times x is x² right?

  26. anonymous
    • 2 years ago
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    Ohhh yeah ! That makes more sense !!

  27. johnweldon1993
    • 2 years ago
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    Alright perfect! so altogether we have x² + 2x - 2x - 4 Anything look like it can be simplified?

  28. anonymous
    • 2 years ago
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    @johnweldon1993 don't leave meee we're so close to finishing thisss D:

  29. johnweldon1993
    • 2 years ago
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    lol no I'm not leaving openstudy is starting to lag on my computer...

  30. johnweldon1993
    • 2 years ago
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    So back to that question... x² + 2x - 2x - 4 anything that can be simplified here?

  31. anonymous
    • 2 years ago
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    2x - 2x ?

  32. johnweldon1993
    • 2 years ago
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    Right! what is 2x - 2x ??

  33. anonymous
    • 2 years ago
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    0 of course :p

  34. johnweldon1993
    • 2 years ago
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    lol making sure you're paying attention :P Okay...so finally we just have x² - 4 Well...this is all we needed...because when we have a constant number in a quadratic...this is the y-intercept... so -4 would be this y-intercept... 2 x-intercepts at x = -2 and x = 2...and 1 y-intercept at -4 so altogether we have a graph that looks like... |dw:1386884802676:dw|

  35. anonymous
    • 2 years ago
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    And that's it ?? :o

  36. johnweldon1993
    • 2 years ago
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    Thats it :) much easier to put it in a graphing calculator! lol

  37. johnweldon1993
    • 2 years ago
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    And it's good because everything we just did...will act as your explanation of what the key features of your graph are...the x and y intercepts :)

  38. anonymous
    • 2 years ago
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    OMG THANK YOU SOO MUCH !! :DD

  39. johnweldon1993
    • 2 years ago
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    Anytime :) I hope it all made sense!

  40. johnweldon1993
    • 2 years ago
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    And if you ever need help with anything feel free to tag me in a question or message me :)

  41. anonymous
    • 2 years ago
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    I definitely will ! :)

  42. anonymous
    • 2 years ago
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    I actually need help with a question following this one !! :o @johnweldon1993

  43. anonymous
    • 2 years ago
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    Using your graph from question 4, describe if the average rate of change is increasing or decreasing, from left to right. Justify your observations with calculations.

  44. johnweldon1993
    • 2 years ago
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    Oh sorry, didn't give me the notification you replied here... Well lets see...from (-infinity to 0) the function is decreasing....and from (0 to +infinity) the function is increasing...

  45. anonymous
    • 2 years ago
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    Right :o

  46. johnweldon1993
    • 2 years ago
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    For the justification part... \[\large \frac{f(x_2) - f(x_1)}{x_2 - x_1}\] so say that from x = -4 to x = -1 we have f(x) = x² - 4 remember? so plug in x = -4 and x = -1 to that to get f(x2) and f(x1) f(x2) = -1² - 4 = -3 f(x1) = -4² - 4 = 12 so \[\large \frac{-3 - 12}{-1 + 4}\] \[\large \frac{-15}{3}\] \[\large -5 \] since this is a negative number we have a decreasing interval... make sense?

  47. anonymous
    • 2 years ago
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    Makes sense !

  48. johnweldon1993
    • 2 years ago
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    Perfect :)

  49. anonymous
    • 2 years ago
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    So that's it ?? :o lol i'm a little slow

  50. johnweldon1993
    • 2 years ago
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    Yeah that's it...if you wanted to do another calculation for the positive side you could but....yeah lol...and you're not slow you got exactly what I was saying :)

  51. anonymous
    • 2 years ago
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    Thank you againnn :)

  52. johnweldon1993
    • 2 years ago
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    Of course :)

  53. anonymous
    • 9 months ago
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    im having trouble with a similar problem, (x+3)(x-3) x = 3 y = -9 I have graphed. i just dont understand : describe if the average rate of change is increasing or decreasing, from left to right. Justify your observations with calculations. like how do i do/start this?

  54. anonymous
    • 9 months ago
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    @mathmate @abb0t @ganeshie8 @SithsAndGiggles @zepdrix @Compassionate much appreciation for help.

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