Ros is trying to find the solution(s) to the system {f(x)=βx3+2x2+xβ2g(x)=x2βxβ2.
roz wan...
Mathematics, 25.01.2020 12:31, marialion5278
Ros is trying to find the solution(s) to the system {f(x)=βx3+2x2+xβ2g(x)=x2βxβ2.
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,β2), (2,0), and (β1,0).
which work correctly verifies whether each of rozβs 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(β2) and f(β2)=g(0); f(2)=g(0) and f(0)=g(2), or f(β1)=g(0) and f(0)=g(β1).
1. f(0)=β03+2(02)+0β2=β2; g(β2)=(β2)2β2β2=0 thus, (0,β2) is a solution.
2. f(2)=β23+2(22)+2β2=0; g(0)=02β0β2=2 thus, (2,0) is a solution.
3. f(β1)=β(β1)3+2(β1)2+(β1)β2=0; g(0)=02β0β2=2 thus, (β1,0) 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)=β2, f(2)=g(2)=0, and f(β1)=g(β1)=0 as follows:
1. f(0)=β03+2(02)+0β2=β2; g(0)=02β0β2=β2 thus, (0,β2) is a solution.
2. f(2)=β23+2(22)+2β2=0; g(2)=22β2β2=0 thus, (2,0) is a solution.
3. f(β1)=β(β1)3+2(β1)2+(β1)β2=0; g(β1)=(β1)2β(β1)β2=0 thus, (β1,0) 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(β2)=g(β2)=0, and f(0)=g(0)=2 or f(0)=g(0)=β1 as follows:
1. f(β2)=β23+2(22)+2β2=0; g(β2)=(β2)2β2β2=0 thus, (0,β2) is a solution.
2. f(0)=β03+2(02)β0+2=2; g(0)=02β0β2=2 thus, (2,0) is a solution.
3. since f(0)=g(0)=2, f(0) and g(0) cannot equal β1. thus, (β1,0) 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(β2)=g(β2)=0, and f(0)=g(0)=2 or f(0)=g(0)=β1 as follows:
1. f(β2)=β23+2(22)+2β2=0; g(β2)=(β2)2β2β2=0 thus, (0,β2) is a solution.
2. f(0)=β03+2(02)β0+2=2; g(0)=02β0β2=2 thus, (2,0) is a solution.
since f(0)=g(0)=2, f(0) and g(0) cannot equal β1. thus, (β1,0) is not a solution.
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