first off, let's notice the graph touches the x-axis at -1 and 3, namely, those are the zeros/solutions/roots of the polynomial and therefore, the factors come from those points.
now, at -1, the graph doesn't cross the x-axis, instead it <u>simply bounces off</u> of it, that means the zero of x = -1, has an even multiplicity, could be 4 or 2 or 6, but let's go with 2.
at x = 3, the graph does cross the x-axis, meaning it has an odd multiplicity, could be 3 or 1, or 7 or 9, but let's use 1.
![\bf \begin{cases} x=-1\implies &x+1=0\\ x=3\implies &x-3=0 \end{cases}~\hspace{5em}\stackrel{\textit{even multiplicity}}{(x+1)^2}\qquad \stackrel{\textit{odd multiplicity}}{(x-3)^1}=\stackrel{y}{0} \\\\\\ (x^2+2x+1)(x-3)=y\implies x^3+2x^2+x-3x^2-6x-3=y \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill x^3-x^2-5x-3=y~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%3D-1%5Cimplies%20%26x%2B1%3D0%5C%5C%20x%3D3%5Cimplies%20%26x-3%3D0%20%5Cend%7Bcases%7D~%5Chspace%7B5em%7D%5Cstackrel%7B%5Ctextit%7Beven%20multiplicity%7D%7D%7B%28x%2B1%29%5E2%7D%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bodd%20multiplicity%7D%7D%7B%28x-3%29%5E1%7D%3D%5Cstackrel%7By%7D%7B0%7D%20%5C%5C%5C%5C%5C%5C%20%28x%5E2%2B2x%2B1%29%28x-3%29%3Dy%5Cimplies%20x%5E3%2B2x%5E2%2Bx-3x%5E2-6x-3%3Dy%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20x%5E3-x%5E2-5x-3%3Dy~%5Chfill)
The easiest way to do this is to plug in the numbers for the variebles and see if they equal the same in both sides. lets try the first one 5(1)+2(-3)=-1, multiply the nmbers to get 5-6=-1 now simplify to get the answer of -1=-1, they both equal the same so this means that the first option is the correct one
Hope this helps
Answer:
D
Step-by-step explanation:
The correct answer is <span>C:Greater hours worked, fewer hours spent talking on the phone.
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Answer:
(A) It would have to lie on AB because DC would rotate halfway around M. B would lie on D and A would lie on C.
(B) AD would lie on the BC line. D would lie on B and A would lie on C.
(C) The Lines CD and AD would only intersect if one line had to rotate 180 degrees around M.
