Question

The functions f(x) and g(x) are shown on the graph.

Answer 1

Answer:

$g(x)=\log_3(x+4)$

Step-by-step explanation:

Translations

For $a > 0$

$f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}$

$f(x-a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units right}$

$f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}$

$f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}$

Parent function:

  $f(x)=\log_3x$

From inspection of the graph:

• The x-intercept of $f(x)$ is (1, 0)
• The x-intercept of $g(x)$ 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:

$g(x)=f(x+4)=\log_3(x+4)$

Further Proof

Logs of zero or negative numbers are undefined.

From inspection of the graph, $x=-3$ is part of the domain of g(x).

Therefore, input this value of x into the answer options:

 $g(-3)=\log_3(-3)-4\implies$ undefined

 $g(-3)=\log_3(-3)+4\implies$ undefined

 $g(-3)=\log_3(-3-4)=\log_3(-7)=\implies$ undefined

 $g(-3)=\log_3(-3+4)=\log_3(1)=0$

Hence proving that $g(x)=\log_3(x+4)$

Answer 2

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