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 (–∞, –3) F(x) > 0 over the interval (–∞, –4)

f(x) > 0 over the interval

Step-by-step explanation:

If  f(x)  is a continuous function,  and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.

Then the function was above the x axis (larger than zero) from , until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.

There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from to -3, since between the  values -4 and -3 the function adopts negative values.

answer is 12345678900987654321234567890987654321

step-by-step explanation: