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Math Help - Matching the functons with their graphs

  1. #1
    s3a
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    Matching the functons with their graphs

    How do I do this? If someone could at least show me how to do one of them with full solutions, it would help me understand them all.

    Any help would be greatly appreciated!
    Thanks in advance!
    Attached Thumbnails Attached Thumbnails Matching the functons with their graphs-0001-question-.jpg  
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    wouldnt just using a graping calculator and inputing it in make life sooo much easier?
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    answer to number 10 would be
    1 D
    2 C
    3 B
    4 A
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    Quote Originally Posted by s3a View Post
    How do I do this? If someone could at least show me how to do one of them with full solutions, it would help me understand them all.

    Any help would be greatly appreciated!
    Thanks in advance!
    What you SHOULD be doing is showing us your work, so that we can show you where you're going wrong.

    Having said that...

    7) 3x^4 - 3x^3 - 1x^2 = 0

    x^2(3x^2 - 3x - 1) = 0

    So x^2 = 0 or 3x^2 - 3x - 1 = 0

    Therefore

    x = 0 or x = \frac{3 \pm \sqrt{(-3)^2 - 4(3)(-1)}}{2(3)}

    x = 0 or x = \frac{3 \pm \sqrt{13}}{6}.


    So A = \frac{3 - \sqrt{13}}{6}, B = 0, C = \frac{3 + \sqrt{13}}{6}.
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    Quote Originally Posted by s3a View Post
    How do I do this? If someone could at least show me how to do one of them with full solutions, it would help me understand them all.

    Any help would be greatly appreciated!
    Thanks in advance!
    8) 5 - x < -1

    5 + 1 < x

    x > 6.


    Using interval notation, this is written as x \in (6, \infty).


    9) x^3 - 16x \leq 0

    The easiest way to answer this would be to analyse the graph of y = x^3 - 16x. See where the x-intercepts are and where the graph is negative.
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  6. #6
    s3a
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    Actually, I needed detailed solutions for number 10 (or one of the graphs), sorry I didn't make that clear; I was in a rush.
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    Quote Originally Posted by saar4ever View Post
    wouldnt just using a graping calculator and inputing it in make life sooo much easier?
    Of course it would. And having the answers ahead of time would save the trouble of using a calculator! You wouldn't learn anything in either case, but that's not important is it?
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    Quote Originally Posted by s3a View Post
    Actually, I needed detailed solutions for number 10 (or one of the graphs), sorry I didn't make that clear; I was in a rush.
    Well, what do you KNOW about functions? Can I presume that you know how to factor?

    Function (1), f(x)= -x^2(x^2- 4)= -x^2(x-2)(x+2) obviously has zeros at x= 0, 2, and -2 and only at those points. Only graphs A and D have that property. For x> 2, all factors, x, x-2, and x+2, are positive so, because of the "-" in front, the f(x)< 0 for x> 2 and only D has that property.

    Function (2), f(x)= -x^5+ 5x^3- 4x= -x(x^4- 5x^2+ 4) = -x(x^2- 4)(x^2- 1)= -x(x-2)(x+2)(x-1)(x+1) has zeros at x= 0, x= 2, x= -2, x= 1, and x= -1. C is the only graph that has that property.

    Function (3), f(x)= -x^3+ 2x^2= -x^2(x- 2) has zeros only at x= 0 and x= 2 (and a double root at x= 0 so the graph is tangent to the x-axis at x= 0). B is the only graph having that property.

    Function (4), f(x)= \frac{x^6}{2}- 2x^4=\frac{1}{2}x^4(x^2- 4)= \frac{1}{2}x^4(x-2)(x+2) has zeros at x= 0, x= 2, and x= -2 and, as with (1), A and D have that property. But now, for x> 2, f(x)> 0 and only A has that property.
    Last edited by HallsofIvy; August 25th 2009 at 11:19 AM.
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  9. #9
    s3a
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    I was reading and in the beginning I noticed that (x+2) or (x-2) is negative (in my work - as opposed to positive in yours) (can't tell which one is negative and which one is positive).
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