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Saylilbaby

  • one year ago

help i will also give a medal...

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  1. saylilbaby
    • one year ago
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  2. saylilbaby
    • one year ago
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  3. anonymous
    • one year ago
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    CONFUSING LOOKING LOL

  4. rvc
    • one year ago
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    @Vincent-Lyon.Fr @freckles

  5. Australopithecus
    • one year ago
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    This is actually really easy Note that you can factor some polynomials. What you have to do is simply look at the y intercepts in the graph. given an example the following graph: |dw:1435336071585:dw| NOTE this graph has the point in it (1, -6)

  6. Australopithecus
    • one year ago
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    Note point notation (x,y) 1. List out the y-intercepts. In this example we have y-intercepts: (-1,0), (0,0), and (2,0) 2. Look at this like a factoring problem which is essentially setting up a polynomial so all its y-intercepts are evident. We can deduce that the polynomial has the form based on its y-intercept (x+1)(x-2)x note this agrees with our graph! when, x = -1 ((-1)+1)(x-2)x = 0(x-2)x = 0 when, x = 0 (0)(x+1)(x-2) = 0 when, x = 2 ((2)-2)(x+1)x = 0(x+1)x = 0 3. Lastly we need to check to see if the point present in the graph is present in the function we just created to model the graph. THe point given is (1, -6) plug x = 1 into our function (x+1)(x-2)x (1+1)(1-2)*1 = 2*-1*1 = -2 Notice that what we get is not correct!! Our function doesnt seem to be modeling our graph correctly! DOn't worry this can be fixed easily by using coefficients! What coefficient multiplied by -2 will give -6? the answer is 3 so just multiply 3 by our function 3(x+1)(x-2)x now when x = 1 3(-2) = -6 we have modeled the graph

  7. Australopithecus
    • one year ago
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    Let me know if you follow, I dont think I could explain it any clearer

  8. anonymous
    • one year ago
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    thata=s good enough

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