Question

$$g(x) = \
LATEXBLOCK0
$$

what is the graph of $g$? choose 1 answer: a graph a b graph b

Explanation:

Step1: Analyze the first piece of the function

For \( g(x) = -3x - 8 \) with \( -4 \leq x \leq 0 \):

• When \( x = -4 \), \( g(-4)=-3(-4)-8 = 12 - 8 = 4 \). So the point is \( (-4, 4) \).
• When \( x = 0 \), \( g(0)=-3(0)-8=-8 \). So the point is \( (0, -8) \).

Step2: Analyze the second piece of the function

For \( g(x) = -3x + 9 \) with \( 0 < x \leq 3 \):

• When \( x = 0 \) (approaching from the right), \( g(0^+)=-3(0)+9 = 9 \). But at \( x = 0 \), the first piece gives \( -8 \), so there's a jump at \( x = 0 \).
• When \( x = 3 \), \( g(3)=-3(3)+9 = 0 \). So the point is \( (3, 0) \).

Now, check the graphs:

• Graph A: At \( x = 2 \) (in \( 0 < x \leq 3 \)), the y - value should be \( -3(2)+9 = 3 \), but the graph in A at \( x = 2 \) has a different y - value.
• Graph B: For the first piece, when \( x=-4 \), \( y = 4 \) (matches), at \( x = 0 \), \( y=-8 \) (matches). For the second piece, at \( x = 3 \), \( y = 0 \) (matches), and the slope of both pieces is - 3 (since the coefficient of x is - 3 in both linear functions), so the lines are parallel (same slope) with a jump at \( x = 0 \) from \( y=-8 \) (first piece) to \( y = 9 \) (second piece at \( x = 0^+ \)).

Answer:

B