MTH1010_04_04_2

Question

MTH1010_04_04_2

BlackTom AI · Accepted Answer

We start by restating what the graph shows: a straight line with a positive slope, labeled y = f(x), crossing the y-axis at a value around 2 and rising steeply as x increases.
Option (a): y = 2 + 4x. This has a y-intercept of 2 (the value when x = 0) and a slope of 4, meaning the line should rise by 4 units for every 1 unit increase in x. If you plug in x = 0, you get y = 2, which matches the intercept seen on the graph. For x = 1, y = 6, which aligns with the line reaching near y = 6 at x = 1 on the graph. This option is consistent with a steep positive slope and the intercept observed.
Option (b): y = 2 − 4x. This has a negative slope (−4), so the line would slope downward as x increases. The graph clearly shows an upward sloping line, so this choice contradicts the visual trend of the line and is incorrect.
Option (c): y = 4 + 2x. This has a y-intercept of 4 and a smaller positive slope of 2. While it yields an increasing line, the intercept would be at y = 4, which does not match the line’s apparent intercept around y = 2 on the graph. Additionally, the slope would be more gradual than what the graph exhibits, making this option unlikely.
Option (d): y = 4 − 2x. This has a negative slope (−2) and an intercept of 4. The negative slope contradicts the graph’s upward trend, and the intercept value does not align with the observed crossing at y ≈ 2. Hence this is not a match.
In summary, the key features to compare are the sign of the slope (positive in the graph) and the y-intercept value (close to 2). Among the options, only y = 2 + 4x satisfies both: a positive slope of 4 and a y-intercept of 2, matching the graph’s behavior and crossing points.

MTH1010_04_04_2

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