Question

Which function is positive for the entire interval [–3, –2]? On a coordinate plane, a curved line with a minimum value of (0, negative 3) crosses the x-axis at (negative 3, 0) and (3, 0), and crosses the y-axis at (0, negative 3). On a coordinate plane, a curved line with a minimum value of (2, negative 3) crosses the x-axis at (negative 1, 0) and (5, 0), and crosses the y-axis at (0, negative 1.5). On a coordinate plane, a curved line with a minimum value of (2, 4) and a maximum value of (0.5, 6), crosses the x-axis at (negative 1.5, 0) and crosses the y-axis at (0, 5). On a coordinate plane, a curved line with a minimum value of (negative 1.75, negative 3.9) and a maximum value of (0, 2), crosses the x-axis at (negative 2.2, 0), (negative 0.75, 0), and (0.75, 0), and crosses the y-axis at (0, 2).

Answer 1

Answer:

The answer should be the second option!

Step-by-step explanation:

Hope This Helps

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~Zero~

Answer 2

Answer: D

f(x) > 0 over the interval

Step-by-step explanation:

If f(x) is a continuous function and all of the critical points of behavior change are described by the given information, we can say that the function crossed the x axis to reach a minimum value of -12 at 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 is thus most likely a local maximum.

The function was then above the x axis (greater than zero) from till it crossed the x axis (becoming negative) at x = -4. As a result, the function was positive (greater than zero) in this period.

There is no such thing as a unique declaration about the function's positive or negative value when The interval is extended from -4 to -3, because the function takes negative values between -4 and -3.