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Mathematics, 25.01.2020 12:31 marialion5278

Ros is trying to find the solution(s) to the system {f(x)=βˆ’x3+2x2+xβˆ’2g(x)=x2βˆ’xβˆ’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,βˆ’2), (2,0), and (βˆ’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(βˆ’2) and f(βˆ’2)=g(0); f(2)=g(0) and f(0)=g(2), or f(βˆ’1)=g(0) and f(0)=g(βˆ’1).

1. f(0)=βˆ’03+2(02)+0βˆ’2=βˆ’2; g(βˆ’2)=(βˆ’2)2βˆ’2βˆ’2=0 thus, (0,βˆ’2) is a solution.

2. f(2)=βˆ’23+2(22)+2βˆ’2=0; g(0)=02βˆ’0βˆ’2=2 thus, (2,0) is a solution.

3. f(βˆ’1)=βˆ’(βˆ’1)3+2(βˆ’1)2+(βˆ’1)βˆ’2=0; g(0)=02βˆ’0βˆ’2=2 thus, (βˆ’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)=βˆ’2, f(2)=g(2)=0, and f(βˆ’1)=g(βˆ’1)=0 as follows:

1. f(0)=βˆ’03+2(02)+0βˆ’2=βˆ’2; g(0)=02βˆ’0βˆ’2=βˆ’2 thus, (0,βˆ’2) is a solution.

2. f(2)=βˆ’23+2(22)+2βˆ’2=0; g(2)=22βˆ’2βˆ’2=0 thus, (2,0) is a solution.

3. f(βˆ’1)=βˆ’(βˆ’1)3+2(βˆ’1)2+(βˆ’1)βˆ’2=0; g(βˆ’1)=(βˆ’1)2βˆ’(βˆ’1)βˆ’2=0 thus, (βˆ’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(βˆ’2)=g(βˆ’2)=0, and f(0)=g(0)=2 or f(0)=g(0)=βˆ’1 as follows:

1. f(βˆ’2)=βˆ’23+2(22)+2βˆ’2=0; g(βˆ’2)=(βˆ’2)2βˆ’2βˆ’2=0 thus, (0,βˆ’2) is a solution.

2. f(0)=βˆ’03+2(02)βˆ’0+2=2; g(0)=02βˆ’0βˆ’2=2 thus, (2,0) is a solution.

3. since f(0)=g(0)=2, f(0) and g(0) cannot equal βˆ’1. thus, (βˆ’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(βˆ’2)=g(βˆ’2)=0, and f(0)=g(0)=2 or f(0)=g(0)=βˆ’1 as follows:

1. f(βˆ’2)=βˆ’23+2(22)+2βˆ’2=0; g(βˆ’2)=(βˆ’2)2βˆ’2βˆ’2=0 thus, (0,βˆ’2) is a solution.

2. f(0)=βˆ’03+2(02)βˆ’0+2=2; g(0)=02βˆ’0βˆ’2=2 thus, (2,0) is a solution.

since f(0)=g(0)=2, f(0) and g(0) cannot equal βˆ’1. thus, (βˆ’1,0) is not a solution.

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Ros is trying to find the solution(s) to the system {f(x)=βˆ’x3+2x2+xβˆ’2g(x)=x2βˆ’xβˆ’2.

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