Question

which statement is true regarding the functions on the graph?\
\\( \bigcirc \\ f(6) = g(3) \\)\
\\( \bigcirc \\ f(3) = g(3) \\)\
\\( \bigcirc \\ f(3) = g(6) \\)\
\\( \bigcirc \\ f(6) = g(6) \\)

Explanation:

Step1: Analyze \( g(x) \) (red line)

Find \( g(3) \) and \( g(6) \). From the graph, \( g(x) \) has a slope. Let's find its equation. The y-intercept is 3? Wait, no, when \( x = -4 \), \( g(x)=0 \), and when \( x = 0 \), \( g(x)=3 \)? Wait, no, looking at the grid: when \( x = 0 \), \( g(0)=3 \)? Wait, the red line passes through \( (-4, 0) \) and \( (0, 3) \)? Wait, no, the red line (g(x)): when \( x = 3 \), let's see the y-value. From the graph, at \( x = 3 \), the red line (g(x)) is at y = 6? Wait, no, the blue line is f(x). Wait, let's re-examine:

Wait, the red line (g(x)): when \( x = 0 \), y = 3? Wait, no, the grid: each square is 1 unit. Let's find the equation of \( g(x) \). The red line passes through \( (-4, 0) \) and \( (0, 3) \)? Wait, no, when \( x = -4 \), y = 0; when \( x = 0 \), y = 3? Wait, slope \( m = \frac{3 - 0}{0 - (-4)} = \frac{3}{4} \)? No, that doesn't match. Wait, maybe better to read the points. At \( x = 3 \), the red line (g(x)): looking at the graph, when x=3, the red line is at y=6? Wait, the blue line (f(x)): when x=1, y=0? Wait, no, the blue line (f(x)): when x=0, y=-4? Wait, no, the blue line: when x=1, y=0? Wait, let's find the equations.

For \( g(x) \) (red line): Let's take two points. When \( x = -4 \), \( g(-4) = 0 \); when \( x = 0 \), \( g(0) = 3 \). So slope \( m = \frac{3 - 0}{0 - (-4)} = \frac{3}{4} \). So equation: \( g(x) = \frac{3}{4}x + 3 \). Wait, when \( x = 3 \), \( g(3) = \frac{3}{4}(3) + 3 = \frac{9}{4} + 3 = \frac{21}{4} = 5.25 \)? No, that doesn't match the graph. Wait, maybe I misread the points. Wait, the red line (g(x)): when x=3, the y-value is 6? Wait, the blue line (f(x)): when x=3, y=6? Wait, no, the blue line is steeper. Wait, maybe the red line (g(x)) has a slope of 1? Wait, when x=0, y=3; x=3, y=6. So slope 1. So equation \( g(x) = x + 3 \). Let's check: when x=-4, y=-4 + 3 = -1? No, that's not 0. Wait, maybe the red line passes through (-4, 0) and (3, 6). Then slope is \( \frac{6 - 0}{3 - (-4)} = \frac{6}{7} \), no. This is confusing. Maybe better to read the y-values at specific x.

For \( f(x) \) (blue line): Let's find its equation. When x=1, y=0; x=0, y=-4? Wait, no, when x=0, y=-4? Wait, the blue line crosses the y-axis at (0, -4)? Then when x=1, y=0: slope is \( \frac{0 - (-4)}{1 - 0} = 4 \). So equation: \( f(x) = 4x - 4 \). Let's check: when x=1, 4(1)-4=0, correct. When x=3, \( f(3) = 4(3) - 4 = 8 \)? No, that doesn't match the graph. Wait, the graph shows that at x=3, both lines meet? Wait, the intersection point is at (3, 6)? So when x=3, both f(3) and g(3) are 6? Wait, no, the red line and blue line intersect at (3, 6)? So f(3) = 6 and g(3) = 6? Wait, no, the options: f(3)=g(3) is an option. Wait, let's check the options:

Option 1: f(6) = g(3). Let's find f(6): using f(x) = 4x -4, f(6)=24-4=20. g(3): if g(x) at x=3 is 6, then 20≠6. No.

Option 2: f(3)=g(3). If intersection at (3,6), then f(3)=6 and g(3)=6. So this would be true.

Option 3: f(3)=g(6). f(3)=6. g(6): if g(x) is x+3, then g(6)=9. 6≠9. No.

Option 4: f(6)=g(6). f(6)=20, g(6)=9. No.

Wait, maybe the intersection point is (3,6), so f(3)=g(3)=6. So the correct option is f(3)=g(3).

Wait, let's re-express:

From the graph, the two lines intersect at x=3, so at x=3, both functions have the same y-value. So f(3) = g(3).

Step2: Verify each option

- Option 1: f(6) = g(3). Calculate f(6) and g(3). From graph, f(6) is higher (steeper line), g(3) is 6 (if intersection at 3,6). f(6) would be 4*6 -4=20, g(3)=6. Not equal.
- Option 2: f(3)=g(3). At x=3, both lines meet, so y-values are equal. So this is true.
- Option 3: f(3)=g(6). f(3)=6, g(6) would be higher (since g(x) has positive slope), so 6≠g(6).
- Option 4: f(6)=g(6). f(6) is much higher, so no.

Answer:

B. \( f(3) = g(3) \) (assuming the options are labeled as A: \( f(6) = g(3) \), B: \( f(3) = g(3) \), C: \( f(3) = g(6) \), D: \( f(6) = g(6) \))