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Mathematics, 23.03.2020 08:43 mmaglaya1

Ros is trying to find the solution(s) to the system {f(x)=−x3+2x2+x−2g(x)=x2−x−2.

Roz wants to find the solution(s) to this system. After analyzing the graph of the functions, Roz comes up with the following list ordered pairs as possible solutions: (0,−2), (2,0), and (−1,0).

Which work correctly verifies whether each of Roz’s ordered pairs is a solution?

a. A solution to the system will be the intersection of f(x) and g(x) such that f(x)=g(y). Roz must verify one of the following: f(0)=g(−2) and f(−2)=g(0); f(2)=g(0) and f(0)=g(2), or f(−1)=g(0) and f(0)=g(−1).
f(0)=−03+2(02)+0−2=−2; g(−2)=(−2)2−2−2=0 Thus, (0,−2) is a solution.
f(2)=−23+2(22)+2−2=0; g(0)=02−0−2=2 Thus, (2,0) is a solution.
f(−1)=−(−1)3+2(−1)2+(−1)−2=0; g(0)=02−0−2=2 Thus, (−1,0) is not a solution.

b. A solution to the system will be the intersection of f(x) and g(x) such that f(x)=g(x). Roz must verify that f(0)=g(0)=−2, f(2)=g(2)=0, and f(−1)=g(−1)=0 as follows:
f(0)=−03+2(02)+0−2=−2; g(0)=02−0−2=−2 Thus, (0,−2) is a solution.
f(2)=−23+2(22)+2−2=0; g(2)=22−2−2=0 Thus, (2,0) is a solution.
f(−1)=−(−1)3+2(−1)2+(−1)−2=0; g(−1)=(−1)2−(−1)−2=0 Thus, (−1,0) is a solution.

c. A solution to the system will be the intersection of f(x) and g(x) such that f(x)=g(x). Roz must verify that f(−2)=g(−2)=0, and f(0)=g(0)=2 or f(0)=g(0)=−1 as follows:
f(−2)=−23+2(22)+2−2=0; g(−2)=(−2)2−2−2=0 Thus, (0,−2) is a solution.
f(0)=−03+2(02)−0+2=2; g(0)=02−0−2=2 Thus, (2,0) is a solution.
Since f(0)=g(0)=2, f(0) and g(0) cannot equal −1. Thus, (−1,0) is not a solution.

d. A solution to the system will be the intersection of f(x) and g(x) such that f(x)=g(x). Roz must verify that f(0)=g(0)=−2, f(2)=g(2)=0, and f(−1)=g(−1)=0 as follows:
f(−2)=−23+2(22)+2−2=16; g(−2)=(−2)2−(−2)−2=4 Thus, (0,−2) is not a solution.
f(0)=03+2(02)+0−2=−2; g(0)=02−0−2=−2. Thus, (2,0) is not a solution.
Since f(0)=g(0)=2, f(0) and g(0) cannot equal −1. Thus, (−1,0) is not a solution.

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Ros is trying to find the solution(s) to the system {f(x)=−x3+2x2+x−2g(x)=x2−x−2.

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