anonymous
  • anonymous
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
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@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?
anonymous
  • anonymous
Loser66
  • Loser66
the cube function ( exponent 3) with positive leading coefficient has the left end at \(-\infty\) and right end at \(+ \infty\)|dw:1439221941349:dw|

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Loser66
  • Loser66
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
anonymous
  • anonymous
so how would i put it into words though
Loser66
  • Loser66
Put as what I wrote above.
anonymous
  • anonymous
ohh ok
Loser66
  • Loser66
\(f(x) = x^2(x+2)=x^3+2x^2\) The leading coefficient is 1 >0 , hence the graph will start at -inf and end at + inf Dat sit
anonymous
  • anonymous
1>0
anonymous
  • anonymous
brb
Loser66
  • Loser66
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.
Loser66
  • Loser66
y-intercepts happen when x =0, replace x =0 to solve for y

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