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A teacher gave her class two functions, f(x) = |x – 5|+4 and g(x) = -x² + 21, and aske for the solution(s) of the equation f(x) = g(x). The answers given by four students are shown below: Betty: only 4 Donna: only 5 Edward: 12 and 5 Gary: -3 and 4 Who found the solution correctly and why? A. Betty, because the solutions are the x-values of the points of intersection of the graphs of f(x) and g(x) but the negative solutions must be discarded because f(x) is an absolute value function. B. Edward, the solutions are the y-values of the points of intersection of the graphs of f(x) and g(x). C. Donna, the solutions are the y-values of the points of intersection of the graphs of f(x) and g(x) but solutions paired with negative x-values must be discarded because f(x) is an absolute value function. D. Gary, the solutions are the x-values of the points of intersection of the graphs of

A teacher gave her class two functions, f(x) = |x – 5|+4 and g(x) = -x² + 21, and aske for the solution(s) of the equation f(x) = g(x). The answers given by four students are shown below: Betty: only 4 Donna: only 5 Edward: 12 and 5 Gary: -3 and 4 Who found the solution correctly and why? A. Betty, because the solutions are the x-values of the points of intersection of the graphs of f(x) and g(x) but the negative solutions must be discarded because f(x) is an absolute value function. B. Edward, the solutions are the y-values of the points of intersection of the graphs of f(x) and g(x). C. Donna, the solutions are the y-values of the points of intersection of the graphs of f(x) and g(x) but solutions paired with negative x-values must be discarded because f(x) is an absolute value function. D. Gary, the solutions are the x-values of the points of intersection of the graphs of

Intermediate Algebra
Intermediate Algebra
10th Edition
ISBN:9781285195728
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter7: Equations And Inequalities In Two Variables
Section7.1: Rectangular Coordinate System And Linear Equations
Problem 46PS
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6 of A teacher gave her class two functions, f(x) = |x – 5+4 and g(x) = -x² + 21, and asked for the solution(s) of the equation f(x) = g(x). The answers given by four students are 6. shown below. Betty: only 4 Donna: only 5 Edward: 12 and 5 Gary: -3 and 4 Who found the solution correctly and why? A. Betty, because the solutions are the x-values of the points of intersection of the graphs of f(x) and g(x) but the negative solutions must be discarded because f(x) is an absolute value function. B. Edward, the solutions are the y-values of the points of intersection of the graphs of f(x) and g(x). C. Donna, the solutions are the y-values of the points of intersection of the graphs of f(x) and g(x) but solutions paired with negative x-values must be discarded because f(x) is an absolute value function. D. Gary, the solutions are the x-values of the points of intersection of the graphs of f(x) and g(x).
Transcribed Image Text:6 of A teacher gave her class two functions, f(x) = |x – 5+4 and g(x) = -x² + 21, and asked for the solution(s) of the equation f(x) = g(x). The answers given by four students are 6. shown below. Betty: only 4 Donna: only 5 Edward: 12 and 5 Gary: -3 and 4 Who found the solution correctly and why? A. Betty, because the solutions are the x-values of the points of intersection of the graphs of f(x) and g(x) but the negative solutions must be discarded because f(x) is an absolute value function. B. Edward, the solutions are the y-values of the points of intersection of the graphs of f(x) and g(x). C. Donna, the solutions are the y-values of the points of intersection of the graphs of f(x) and g(x) but solutions paired with negative x-values must be discarded because f(x) is an absolute value function. D. Gary, the solutions are the x-values of the points of intersection of the graphs of f(x) and g(x).
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