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anonymous

  • one year ago

Complete the following: (a)           Use the Leading Coefficient Test to determine the graph's end behavior. (b)           Find the x-intercepts. State whether the graph crosses the x-axis or touches the x-axis and turns around at each intercept. Show your work. (c)           Find the y-intercept. Show your work. f(x) = x2(x + 2) (a).inf (b).touches at 0 and goes through -2 (c).2

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  1. anonymous
    • one year ago
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    @phi my answers are up there but my teacher said this You need to provide the end behavior. On the right, is it going up or down and also what happens on the left? Distribute the x^2 before determining the degree of the polynomial. Then use the Leading Coefficient Test. Refer to section 2-6 for examples and sample problems, here you will find how to describe end behavior. 4b. At the x intercept, y=0. Since this is quadratic, how would you know if it intercepts or touches the x-axis?

  2. anonymous
    • one year ago
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    @Vocaloid

  3. Loser66
    • one year ago
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    the cube function ( exponent 3) with positive leading coefficient has the left end at \(-\infty\) and right end at \(+ \infty\)|dw:1439221941349:dw|

  4. Loser66
    • one year ago
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    Hence, by looking at the leading coefficient ( the number stands right in the front of x^3) , you can see how the end looks like

  5. anonymous
    • one year ago
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    so how would i put it into words though

  6. Loser66
    • one year ago
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    Put as what I wrote above.

  7. anonymous
    • one year ago
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    ohh ok

  8. Loser66
    • one year ago
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    \(f(x) = x^2(x+2)=x^3+2x^2\) The leading coefficient is 1 >0 , hence the graph will start at -inf and end at + inf Dat sit

  9. anonymous
    • one year ago
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    1>0

  10. anonymous
    • one year ago
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    brb

  11. Loser66
    • one year ago
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    to b) x intercepts happen when y =0, that is \(x^2 (x+2) =0\) , solve for x, you will have 2 solutions. Those are x-intercepts.

  12. Loser66
    • one year ago
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    y-intercepts happen when x =0, replace x =0 to solve for y

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