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Sagot :
Answer:
[tex]g(x)=\log_3(x+4)[/tex]
Step-by-step explanation:
Translations
For [tex]a > 0[/tex]
[tex]f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}[/tex]
[tex]f(x-a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units right}[/tex]
[tex]f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}[/tex]
[tex]f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}[/tex]
Parent function:
[tex]f(x)=\log_3x[/tex]
From inspection of the graph:
- The x-intercept of [tex]f(x)[/tex] is (1, 0)
- The x-intercept of [tex]g(x)[/tex] is (-3, 0)
If there was a vertical translation, the end behaviors of both g(x) and f(x) would be the same in that the both functions would be increasing from -∞ in quadrant IV.
As the x-intercepts of both functions is different, and g(x) increases from -∞ in quadrant III, this indicates that there has been a horizontal translation of 4 units to the left.
Therefore:
[tex]g(x)=f(x+4)=\log_3(x+4)[/tex]
Further Proof
Logs of zero or negative numbers are undefined.
From inspection of the graph, [tex]x=-3[/tex] is part of the domain of g(x).
Therefore, input this value of x into the answer options:
[tex]g(-3)=\log_3(-3)-4\implies[/tex] undefined
[tex]g(-3)=\log_3(-3)+4\implies[/tex] undefined
[tex]g(-3)=\log_3(-3-4)=\log_3(-7)=\implies[/tex] undefined
[tex]g(-3)=\log_3(-3+4)=\log_3(1)=0[/tex]
Hence proving that [tex]g(x)=\log_3(x+4)[/tex]
g(x) is the translated edition of f(x)
- f(x)=log_3x
Let y=f(x)
- y=log_3x
Now
The graph is shifted 4 units left means change in x axis.
- y=log_3(x+4)
No change in y
So
- g(x)=log_3(x+4)
Option D
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