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Math Help - Polynomial Functions

  1. #1
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    Polynomial Functions

    Graph each function given below on a graphing calculator to find a general rule for determining when a graph crosses the x axis at an x intercept or when the graph just touches and turns away from the x axis. State the rule that you find. [3]

    y = (x + 1)^2(x - 2)

    y = (x - 4)^3(x - 1)^2

    y = (x - 3)^2(x + 4)^4
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  2. #2
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    Quote Originally Posted by buddah View Post
    Graph each function given below on a graphing calculator to find a general rule for determining when a graph crosses the x axis at an x intercept or when the graph just touches and turns away from the x axis. State the rule that you find. [3]

    y = (x + 1)^2(x - 2)

    y = (x - 4)^3(x - 1)^2

    y = (x - 3)^2(x + 4)^4
    so ... what have you noticed about the zeros of each graph?
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  3. #3
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    there are 3 zeros?
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  4. #4
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    Quote Originally Posted by buddah View Post
    there are 3 zeros?
    Do you know what a "zero" is on a graph?
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  5. #5
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    Here's another guider,

    To find the x-intercepts of a polynomial, we have to make y = 0

    So for your equations

    0 = (x + 1)^2(x - 2)<br /> <br />
0 = (x - 4)^3(x - 1)^2<br /> <br />
0 = (x - 3)^2(x + 4)^4

    Now have a think... How do you make a product of 2 terms 0?
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  6. #6
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    Quote Originally Posted by buddah View Post
    Graph each function given below on a graphing calculator to find a general rule for determining when a graph crosses the x axis at an x intercept or when the graph just touches and turns away from the x axis. State the rule that you find. [3]

    y = (x + 1)^2(x - 2)

    y = (x - 4)^3(x - 1)^2

    y = (x - 3)^2(x + 4)^4
    Surely you can "Graph each function ... on a graphing calculator ..." and make some observations. What have you tried? Where are you stuck?
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Surely you can "Graph each function ... on a graphing calculator ..." and make some observations. What have you tried? Where are you stuck?
    i solved it:

    okay well i tried to do it

    is this correct?

    x intercept y=0

    y = (x + 1)^2(x - 2)
    0 = (x + 1)^2(x - 2)
    (x + 1)^2 = 0 =====> x + 1 = 0 =====> x = -1 (-1 , 0)
    x - 2 = 0 =====> x = 2 (2 , 0)

    y = (x - 4)^3(x - 1)^2
    0 = (x - 4)^3(x - 1)^2
    (x - 4)^3 = 0 ====> x - 4 = 0 =====> x = 4 (4 , 0)
    (x - 1)^2 = 0 ====> x - 1 = 0 ====> x = 1 (1 , 0)

    y = (x - 3)^2(x + 4)^4
    0 = (x - 3)^2(x + 4)^4
    (x - 3)^2 = 0 ====> x - 3 = 0 ====> x = 3 (3 , 0)
    (x + 4)^4 = 0 ===> x + 4 = 0 ====> x = -4 (-4 , 0)

    however am having trouble explaining it
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  8. #8
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    you've missed the point of the original question ...

    Graph each function given below on a graphing calculator to find a general rule for determining when a graph crosses the x axis at an x intercept or when the graph just touches and turns away from the x axis. State the rule that you find.

    y = (x + 1)^2(x - 2)

    y = (x - 4)^3(x - 1)^2

    y = (x - 3)^2(x + 4)^4
    did you graph each function? does the graph of each function cross the x-axis at each zero (x-intercept), or does it just "touch" the x-axis and turn away?

    which zeros cross? which turn away?
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  9. #9
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    y = (x + 1)^2(x - 2)

    Crosses each intercept

    y = (x - 4)^3(x - 1)^2

    Turns away from the x axis

    y = (x - 3)^2(x + 4)^4

    Turns away from the x axis
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  10. #10
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    Quote Originally Posted by buddah View Post
    y = (x + 1)^2(x - 2)

    Crosses each intercept Mr F says: What? Are you saying the grpah crosses BOTH x-intercepts ....!? There is nothing more we can do for you if you cannot correctly examine a graph that you draw on your calculator (assuming you typed it in correctly, and if you haven't then there's still nothing more we can do).

    y = (x - 4)^3(x - 1)^2

    Turns away from the x axis

    y = (x - 3)^2(x + 4)^4

    Turns away from the x axis
    There comes a time when you are expected to apply the help you have been given and finish the question yourself. That time is now.
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  11. #11
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    mmm.. I'll graph the first one for you



    As you can see, there is a x-intercept at -1 and at 2.

    However, the line bounces off -1 but goes straight through 2... now compare this observation with the equation y = (x + 1)^2(x - 2)

    What's special about the  (x+1) factor?
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  12. #12
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    the square?
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  13. #13
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    That's correct. So because (x+1) is squared. If we look on the graph, it bounces off the x axis, it doesn't touch it.

    Now have a look at the second graph:



    This graph is y = (x - 4)^3(x - 1)^2

    You already worked out that there's x-intercepts at 4 and 1. Do you notice any trends? Since the x-1 is squared is bounces off the graph....

    I'm not sure if you've studied cubics before but a perfect cube is one of the most simple examples of a graph that has what we call an "inflection point". So at x=4 there is an inflection point.

    Can you make any conclusions from this?
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    the cubic part is symmetrical? *cantwatchface
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  15. #15
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    lol, don't be shy...

    Here is a standard cubic graph (you don't have to know this but I'll show you anyway...):



    As you can see the "straight" part in the middle is what we call the "Point of Inflection". This is all you need to know, that when there is a shape like that it is called a point of inflection.

    Now if you compile all the stuff you've learnt from the 2 graphs I have shown you in my previous 2 posts can you come to any conclusions?

    I'll start you off...

    If a factor is (x-1) it will cut the x axis
    If a factor is  (x-1)^2 it will....
    If a factor is  (x-1)^3 it will...

    If you can't then you should probably get your teacher to help you because it will make it easier if you have some one on one time with a real teacher
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