Question

draw the graph of $f(x) = \log_{3}(x - 3) - 1$

question 17
solve the given equation for $x$.
$4^{3x - 3} = 48$

Explanation:

Step1: Identify the parent function

The parent function is \( y = \log_3(x) \). The graph of \( y = \log_3(x) \) has a vertical asymptote at \( x = 0 \) and passes through \( (1, 0) \), \( (3, 1) \), \( (9, 2) \), etc.

Step2: Apply horizontal shift

The function \( f(x) = \log_3(x - 3) - 1 \) has a horizontal shift of 3 units to the right. So the vertical asymptote of the parent function \( x = 0 \) shifts to \( x = 3 \). The point \( (1, 0) \) on the parent function shifts to \( (1 + 3, 0) = (4, 0) \), \( (3, 1) \) shifts to \( (3 + 3, 1) = (6, 1) \), \( (9, 2) \) shifts to \( (9 + 3, 2) = (12, 2) \) (but 12 is outside the given grid, so we'll use other points).

Step3: Apply vertical shift

The function also has a vertical shift of 1 unit down. So the point \( (4, 0) \) from the horizontal shift shifts to \( (4, 0 - 1) = (4, -1) \), \( (6, 1) \) shifts to \( (6, 1 - 1) = (6, 0) \), \( (3 + 1, 0) \) [wait, better to find key points: when \( x - 3 = 1 \) (i.e., \( x = 4 \)), \( f(4) = \log_3(1) - 1 = 0 - 1 = -1 \). When \( x - 3 = 3 \) (i.e., \( x = 6 \)), \( f(6) = \log_3(3) - 1 = 1 - 1 = 0 \). When \( x - 3 = 9 \) (i.e., \( x = 12 \)), \( f(12) = \log_3(9) - 1 = 2 - 1 = 1 \) (but 12 is outside the grid, so we can use \( x = 4 \), \( x = 6 \), and the asymptote \( x = 3 \)).

Step4: Draw the asymptote and key points

1. Draw the vertical asymptote at \( x = 3 \) (a dashed line).
2. Plot the point \( (4, -1) \) (since when \( x = 4 \), \( f(4) = -1 \)).
3. Plot the point \( (6, 0) \) (since when \( x = 6 \), \( f(6) = 0 \)).
4. The graph should approach the vertical asymptote \( x = 3 \) from the right (since the domain of \( \log_3(x - 3) \) is \( x > 3 \)) and pass through the plotted points, increasing slowly (since the base 3 is greater than 1, the logarithm function is increasing).

Answer:

To draw the graph of \( f(x) = \log_3(x - 3) - 1 \):

1. Draw a vertical dashed line at \( x = 3 \) (vertical asymptote).
2. Plot the points \( (4, -1) \) and \( (6, 0) \).
3. Draw a smooth curve approaching \( x = 3 \) from the right, passing through the plotted points, and increasing as \( x \) increases.