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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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 26 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 xaxis?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1the cube function ( exponent 3) with positive leading coefficient has the left end at \(\infty\) and right end at \(+ \infty\)dw:1439221941349:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Hence, 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
 one year ago
Best ResponseYou've already chosen the best response.0so how would i put it into words though

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Put as what I wrote above.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\(f(x) = x^2(x+2)=x^3+2x^2\) The leading coefficient is 1 >0 , hence the graph will start at inf and end at + inf Dat sit

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1to b) x intercepts happen when y =0, that is \(x^2 (x+2) =0\) , solve for x, you will have 2 solutions. Those are xintercepts.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1yintercepts happen when x =0, replace x =0 to solve for y
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