Question

the graph of a logarithmic function is given. distract the function for each graph from the following options. drag each function given above into the area below the appropriate graph, depending on which function is represented by which graph. functions: h(x) = 1 - log₇x, f(x) = -log₇x, i(x) = log₇x, g(x) = log₇(x - 1), g(x) = log₇(-x), n(x) = log₇x - 1

Explanation:

Step1: Analyze \( I(x)=\log_7 x \)

The parent logarithmic function \( y = \log_b x \) (here \( b = 7>1 \)) has a vertical asymptote at \( x = 0 \), passes through \( (1,0) \), and is increasing. So we look for a graph with these properties.

Step2: Analyze \( F(x)=-\log_7 x \)

This is a reflection of \( \log_7 x \) over the \( x \)-axis. So it will be decreasing, still with vertical asymptote \( x = 0 \) and passing through \( (1,0) \).

Step3: Analyze \( H(x)=1 - \log_7 x \)

Rewrite as \( H(x)=-\log_7 x + 1 \). It's a reflection of \( \log_7 x \) over the \( x \)-axis and a vertical shift up by 1. So it's decreasing, vertical asymptote \( x = 0 \), and when \( x = 1 \), \( H(1)=1 - 0 = 1 \), so it passes through \( (1,1) \).

Step4: Analyze \( g(x)=\log_7(x - 1) \)

This is a horizontal shift of \( \log_7 x \) to the right by 1 unit. So the vertical asymptote is at \( x = 1 \), and it passes through \( (2,0) \) (since when \( x = 2 \), \( \log_7(2 - 1)=\log_7 1 = 0 \)) and is increasing.

Step5: Analyze \( G(x)=\log_7(-x) \)

The domain is \( x<0 \) (since \( -x>0\Rightarrow x < 0 \)), vertical asymptote at \( x = 0 \), and it's a reflection of \( \log_7 x \) over the \( y \)-axis, so it's decreasing (because the argument is \( -x \), and as \( x\) becomes more negative, \( -x \) increases, but the function of \( \log_7 \) of a positive increasing argument would be increasing? Wait, no: let \( u=-x \), then \( G(x)=\log_7 u \), \( u=-x \) is decreasing for \( x<0 \), and \( \log_7 u \) is increasing for \( u>0 \). So by the chain rule, \( G(x) \) is decreasing (since the outer function is increasing and the inner function is decreasing). It passes through \( (-1,0) \) (since \( \log_7(-(-1))=\log_7 1 = 0 \)).

Step6: Analyze \( N(x)=\log_7 x - 1 \)

This is a vertical shift down by 1 unit of \( \log_7 x \). So it has vertical asymptote \( x = 0 \), passes through \( (1,-1) \) (since \( \log_7 1 - 1 = 0 - 1=-1 \)) and is increasing.

Now, let's match with the graphs:

• Graph 47: Let's assume it has vertical asymptote \( x = 0 \), passes through \( (1,0) \), increasing → \( I(x)=\log_7 x \)
• Graph 48: Vertical asymptote \( x = 0 \), passes through \( (1,1) \), decreasing → \( H(x)=1 - \log_7 x \)
• Graph 49: Vertical asymptote \( x = 0 \), passes through \( (1,0) \), decreasing → \( F(x)=-\log_7 x \)
• Graph 50: Vertical asymptote \( x = 1 \), passes through \( (2,0) \), increasing → \( g(x)=\log_7(x - 1) \)
• Graph 51: Domain \( x<0 \), vertical asymptote \( x = 0 \), passes through \( (-1,0) \), decreasing → \( G(x)=\log_7(-x) \)
• Graph 52: Vertical asymptote \( x = 0 \), passes through \( (1,-1) \), increasing → \( N(x)=\log_7 x - 1 \)

(Note: Since the exact visual details of the graphs are a bit unclear from the image, the above is a general method based on logarithmic function transformations. The key is to use the properties of vertical asymptotes, key points (like where \( x = 1 \) or shifted points), and increasing/decreasing behavior to match each function to its graph.)

Answer:

(Assuming the graphs are labeled 47 - 52 as per the image)

• \( I(x)=\log_7 x \): Graph 47
• \( F(x)=-\log_7 x \): Graph 49
• \( H(x)=1 - \log_7 x \): Graph 48
• \( g(x)=\log_7(x - 1) \): Graph 50
• \( G(x)=\log_7(-x) \): Graph 51
• \( N(x)=\log_7 x - 1 \): Graph 52

(Adjust based on the actual visual of the graphs. For example, if a graph has vertical asymptote at \( x = 1 \), it's \( g(x)=\log_7(x - 1) \), etc.)