Answer :
Let's graph the logarithmic function:
[tex]\[ g(x) = \frac{1}{4} \log_2(x-4) - 3 \][/tex]
### Step-by-Step Solution
1. Understand the function form:
- The function \( \log_2(x-4) \) implies a domain constraint: \( x - 4 > 0 \), or \( x > 4 \). This means the function is only defined for \( x > 4 \).
- The term \( \frac{1}{4} \) is a vertical compression factor.
- Subtracting 3 at the end translates the graph downward by 3 units.
2. Identify the asymptote:
- The function \( \log_2(x-4) \) has a vertical asymptote where the argument inside the logarithm equals zero, which is at \( x = 4 \).
3. Plot points:
- To gather specific points, choose \( x \)-values greater than 4.
- Let's pick \( x = 8 \) and \( x = 12 \).
4. Calculate \( y \)-values for these points:
- For \( x = 8 \):
[tex]\[ g(8) = \frac{1}{4} \log_2(8-4) - 3 = \frac{1}{4} \log_2(4) - 3 = \frac{1}{4} \cdot 2 - 3 = \frac{1}{2} - 3 = -2.5 \][/tex]
- For \( x = 12 \):
[tex]\[ g(12) = \frac{1}{4} \log_2(12-4) - 3 = \frac{1}{4} \log_2(8) - 3 = \frac{1}{4} \cdot 3 - 3 = \frac{3}{4} - 3 = -2.25 \][/tex]
5. Graph the function:
- Plot the points \( (8, -2.5) \) and \( (12, -2.25) \).
- Draw the vertical asymptote at \( x = 4 \).
### Plot Description:
- Vertical asymptote: Draw a dashed vertical line at \( x = 4 \).
- Points:
- \( (8, -2.5) \) - Plot this point and label it.
- \( (12, -2.25) \) - Plot this point and label it.
- Draw the curve: Sketch the curve starting just to the right of \( x = 4 \) and passing through the plotted points. The curve should approach but never intersect the line \( x = 4 \), and continue to decrease slowly as \( x \) increases.
### Final Graph
The graph displays a logarithmic curve that becomes steeper as x approaches 4 from the right and flattens out as x increases. The curve passes through the points (8, -2.5) and (12, -2.25), with a vertical asymptote at \( x = 4 \).
By following these steps, you can graph the logarithmic function [tex]\( g(x) = \frac{1}{4} \log_2(x-4) - 3 \)[/tex], plot two meaningful points on it, and highlight the asymptote.
[tex]\[ g(x) = \frac{1}{4} \log_2(x-4) - 3 \][/tex]
### Step-by-Step Solution
1. Understand the function form:
- The function \( \log_2(x-4) \) implies a domain constraint: \( x - 4 > 0 \), or \( x > 4 \). This means the function is only defined for \( x > 4 \).
- The term \( \frac{1}{4} \) is a vertical compression factor.
- Subtracting 3 at the end translates the graph downward by 3 units.
2. Identify the asymptote:
- The function \( \log_2(x-4) \) has a vertical asymptote where the argument inside the logarithm equals zero, which is at \( x = 4 \).
3. Plot points:
- To gather specific points, choose \( x \)-values greater than 4.
- Let's pick \( x = 8 \) and \( x = 12 \).
4. Calculate \( y \)-values for these points:
- For \( x = 8 \):
[tex]\[ g(8) = \frac{1}{4} \log_2(8-4) - 3 = \frac{1}{4} \log_2(4) - 3 = \frac{1}{4} \cdot 2 - 3 = \frac{1}{2} - 3 = -2.5 \][/tex]
- For \( x = 12 \):
[tex]\[ g(12) = \frac{1}{4} \log_2(12-4) - 3 = \frac{1}{4} \log_2(8) - 3 = \frac{1}{4} \cdot 3 - 3 = \frac{3}{4} - 3 = -2.25 \][/tex]
5. Graph the function:
- Plot the points \( (8, -2.5) \) and \( (12, -2.25) \).
- Draw the vertical asymptote at \( x = 4 \).
### Plot Description:
- Vertical asymptote: Draw a dashed vertical line at \( x = 4 \).
- Points:
- \( (8, -2.5) \) - Plot this point and label it.
- \( (12, -2.25) \) - Plot this point and label it.
- Draw the curve: Sketch the curve starting just to the right of \( x = 4 \) and passing through the plotted points. The curve should approach but never intersect the line \( x = 4 \), and continue to decrease slowly as \( x \) increases.
### Final Graph
The graph displays a logarithmic curve that becomes steeper as x approaches 4 from the right and flattens out as x increases. The curve passes through the points (8, -2.5) and (12, -2.25), with a vertical asymptote at \( x = 4 \).
By following these steps, you can graph the logarithmic function [tex]\( g(x) = \frac{1}{4} \log_2(x-4) - 3 \)[/tex], plot two meaningful points on it, and highlight the asymptote.