Question

use the interactive graph below to sketch a graph of $y = 4\log_{3}(-x - 3)$. place the asymptote before placing the two points.

Explanation:

Step1: Find the vertical asymptote

For the logarithmic function \( y = a\log_b(u) \), the vertical asymptote occurs where \( u = 0 \). Here, \( u=-x - 3 \), so set \( -x - 3=0 \). Solving for \( x \): \( -x=3 \) gives \( x=-3 \). So the vertical asymptote is \( x = - 3 \).

Step2: Find two points on the graph

Point 1: Let \( -x - 3 = 1 \) (since \( \log_b(1)=0 \) for any \( b>0,b

eq1 \))
Solve \( -x - 3 = 1 \): \( -x=4 \) so \( x=-4 \). Then \( y = 4\log_3(1)=4\times0 = 0 \). So one point is \( (-4,0) \).

Point 2: Let \( -x - 3 = 3 \) (since \( \log_3(3)=1 \))

Solve \( -x - 3 = 3 \): \( -x=6 \) so \( x=-6 \). Then \( y = 4\log_3(3)=4\times1 = 4 \). Wait, but looking at the graph, maybe another approach. Alternatively, let's check when \( x=-4 \), we have \( y = 4\log_3(1)=0 \) (point \( (-4,0) \)). When \( x=-1 \), \( -x - 3=-1 - 3=-4 \), but log of negative is undefined. Wait, maybe I made a mistake. Wait, the function is \( y = 4\log_3(-x - 3) \), so the argument \( -x - 3>0\Rightarrow -x>3\Rightarrow x < - 3 \). Oh! I had a mistake earlier. The domain is \( x < - 3 \), so my previous point with \( x=-4 \) is valid (since \( -4 < - 3 \)), \( x=-6 \) is also valid (\( -6 < - 3 \)). Wait, but the graph in the image has points at \( (1,0) \) and \( (5,5) \)? No, maybe the graph in the image is a different function, but for our function \( y = 4\log_3(-x - 3) \), let's correct.

Wait, let's re - evaluate. The function \( y = 4\log_3(-x - 3) \) can be rewritten as \( y = 4\log_3(-(x + 3)) \), which is a reflection over the y - axis and a horizontal shift left by 3 units of \( y = 4\log_3(x) \).

Let's find points correctly:

Case 1: Let \( -x - 3=1\Rightarrow x=-4 \), \( y = 4\log_3(1)=0 \) (point \( (-4,0) \))

Case 2: Let \( -x - 3 = \frac{1}{3}\Rightarrow -x=3+\frac{1}{3}=\frac{10}{3}\Rightarrow x =-\frac{10}{3}\approx - 3.333 \), \( y = 4\log_3(\frac{1}{3})=4\times(- 1)=-4 \)

Case 3: Let \( -x - 3 = 3\Rightarrow x=-6 \), \( y = 4\log_3(3)=4\times1 = 4 \) (point \( (-6,4) \))

But maybe the intended points are based on the transformation. The parent function \( y=\log_3(x) \) has vertical asymptote \( x = 0 \), point \( (1,0) \), \( (3,1) \). For \( y = 4\log_3(-x - 3) \), we have a reflection over the y - axis (replace \( x \) with \( -x \)) and a horizontal shift left by 3 units (replace \( x \) with \( x + 3 \)). So the vertical asymptote of \( y=\log_3(x) \) is \( x = 0 \), after reflection over y - axis it's \( x = 0 \), then shift left 3 units: \( x=-3 \) (which matches our earlier calculation). The point \( (1,0) \) on \( y=\log_3(x) \), after reflection over y - axis becomes \( (-1,0) \), then shift left 3 units: \( x=-1-3=-4 \), \( y = 0 \) (so point \( (-4,0) \)). The point \( (3,1) \) on \( y=\log_3(x) \), after reflection over y - axis becomes \( (-3,1) \), then shift left 3 units: \( x=-3 - 3=-6 \), \( y = 4\times1 = 4 \) (so point \( (-6,4) \)).

Answer:

The vertical asymptote is \( x=-3 \). Two points on the graph are \( (-4,0) \) and \( (-6,4) \) (or other valid points within the domain \( x < - 3 \) calculated using the function \( y = 4\log_3(-x - 3) \)). To sketch the graph, first draw the vertical asymptote \( x=-3 \), then plot the points \( (-4,0) \) and \( (-6,4) \) (or other valid points) and draw the curve approaching the asymptote as \( x
ightarrow - 3^+ \) (from the right - hand side of the asymptote, since \( x < - 3 \)) and extending to the left.