On a coordinate plane, a curved line with a minimum value of (negative 2.5, negative 12) and a maximum value of (0, negative 3) crosses the x-axis at (negative 4, 0) and crosses the y-axis at (0, negative 3).
Which statement is true about the graphed function?

F(x) < 0 over the interval (β€“βˆž, –4)
F(x) < 0 over the interval (β€“βˆž, –3)
F(x) > 0 over the interval (β€“βˆž, –3)
F(x) > 0 over the interval (β€“βˆž, –4)

Respuesta :

The points on the graph of (-4, 0), (-2.5, -12), and (0, -3), gives;

  • F(x) > 0 over the interval (-∞, -4)

Which method can be used to find the true statement?

From the description of the graph, we have;

Furthest point left of the graph = (-4, 0)

The furthest point right on the graph = (0, -3) = The maximum point

The minimum point = (-2.5, -12)

F(x) < 0 at the minimum point

The minimum point is to the right of x = -4

The point the graph crosses the y-axis = (0, -3)

Therefore;

The interval of the graph where F(x) is larger than 0 is to the left of (-4, 0), is the interval (-∞, -4)

The true statement is therefore;

  • F(x) > 0 over the interval (-∞, -4)

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