Question

1. $g(x) = \frac{1}{3}|x + 2| + 6$

Explanation:

Step1: Recall absolute value function form

The general form of an absolute value function is \( y = a|x - h| + k \), where \((h, k)\) is the vertex, \(a\) determines the vertical stretch/compression and direction. For \( g(x)=\frac{1}{3}|x + 2|+6 \), we can rewrite \( x + 2 \) as \( x-(-2) \), so \( h=-2 \), \( k = 6 \), \( a=\frac{1}{3} \).

Step2: Identify vertex

The vertex of \( g(x) \) is at \((-2, 6)\).

Step3: Analyze vertical stretch

Since \( a=\frac{1}{3}\) (positive, so the graph opens upwards) and \( 0<\frac{1}{3}<1 \), the graph is a vertical compression of the parent function \( y = |x| \).

Step4: Plot key points

• For \( x=-2 \), \( g(-2)=\frac{1}{3}|0| + 6=6 \) (vertex).
• For \( x = 1 \) (3 units to the right of -2), \( g(1)=\frac{1}{3}|3|+6 = 1 + 6=7 \).
• For \( x=-5 \) (3 units to the left of -2), \( g(-5)=\frac{1}{3}| - 3|+6=1 + 6 = 7 \).

We can then sketch the graph using these points, with the vertex at \((-2, 6)\), opening upwards, and a vertical compression factor of \( \frac{1}{3} \).

(Note: Since the problem likely asks to graph the function, the above steps guide the graphing process. If the question was about identifying the graph's features, the vertex is \((-2, 6)\), vertical compression by \( \frac{1}{3} \), opens up.)

Answer:

To graph \( g(x)=\frac{1}{3}|x + 2|+6 \):

1. Vertex: Plot \((-2, 6)\) (from \( y = a|x - h|+k \), \( h=-2, k = 6 \)).
2. Direction/Stretch: Opens upward ( \( a=\frac{1}{3}>0 \) ) with vertical compression ( \( |a|=\frac{1}{3}<1 \) ).
3. Additional Points: Use \( x=-5 \) ( \( g(-5)=7 \) ) and \( x = 1 \) ( \( g(1)=7 \) ) to draw the two arms of the V - shaped graph, passing through \((-5, 7)\), \((-2, 6)\), \((1, 7)\) and symmetric about \( x=-2 \).

(If the task was to identify the graph’s vertex, the answer is vertex at \(\boldsymbol{(-2, 6)}\); if to describe the graph, it’s a vertically compressed (by \( \frac{1}{3} \)), upward - opening absolute - value graph with vertex at \((-2, 6)\).)