A logarithmic curve increasing to the right is graphed on a coordinate grid. The domain of the curve is left-parenthesis 3 comma infinity right-parenthesis. The curve passes through the point left-parenthesis 3 comma 0 right-parenthesis.

Question

Question
Use the graph to answer the question.
Which of the following functions is represented by the graph?
(1 point)
Responses

f left parenthesis x right parenthesis equals log subscript 4 baseline x minus 3
Image with alt text: f left parenthesis x right parenthesis equals log subscript 4 baseline x minus 3

f left parenthesis x right parenthesis equals log subscript 4 baseline x plus 3
Image with alt text: f left parenthesis x right parenthesis equals log subscript 4 baseline x plus 3

f left parenthesis x right parenthesis equals log subscript 4 baseline left parenthesis x minus 3 right parenthesis
Image with alt text: f left parenthesis x right parenthesis equals log subscript 4 baseline left parenthesis x minus 3 right parenthesis

f left parenthesis x right parenthesis equals log subscript 4 baseline left parenthesis x plus 3 right parenthesis

Answer 1

To determine which function is represented by the curve described in your question, we need to consider a few key points:

1. Domain: You mentioned that the domain of the curve is \((3, \infty)\). This means the function must be defined for values greater than 3. This information suggests that the argument inside the logarithm function must also be greater than 0 when \(x > 3\).

2. Point (3, 0): The curve passes through the point (3, 0). This indicates that the function value equals 0 when \(x = 3\). The logarithm of a value equals 0 when that value is equal to 1 (because \(\log_b(1) = 0\) for any base \(b\)). Therefore, we need the argument of the logarithm to equal 1 when \(x = 3\).

Let's analyze the given options:

1. \(f(x) = \log_4(x - 3)\)

   • The domain is \((3, \infty)\) (valid).
   • At \(x = 3\): \(f(3) = \log_4(3 - 3) = \log_4(0)\) (undefined).

2. \(f(x) = \log_4(x + 3)\)

   • The domain is \((-3, \infty)\) (not valid since the question states the domain starts at 3).

3. \(f(x) = \log_4(x - 3)\)

   • The domain is \((3, \infty)\) (valid).
   • At \(x = 3\): \(f(3) = \log_4(3 - 3) = \log_4(0)\) (undefined).

4. \(f(x) = \log_4(x + 3)\)

   • The domain is \((-3, \infty)\) (not valid since the question states the domain starts at 3).

Given the options provided, it seems there has been a repeat in options 1 and 3. However, for a correct option that passes through (3,0) and has a valid domain \((3, \infty)\), we can conclude that it must fit the form:

• \(f(x) = \log_4(x - 3)\)

Since that is the only function whose domain matches and would theoretically pass through the point of interest (if we are treating it as a limiting case, the behavior tells us it starts from zero at the limit of the domain), we can opt for the logarithmic function that matches, yet we keep in mind that none are perfectly stated as they are the same choice.

Thus the best selection would be:

Response: \(f(x) = \log_4(x - 3)\).