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

User AstroBen
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2.7k points

1 Answer

21 votes
21 votes

Answer:

f(x) > 0 over the interval (–∞, –4)

Explanation:

We know that:

Our curve has a minimum at (-2.5, -12)

Our curve has a maximum at (0, - 3) (I assume that is a local maximum).

We know that the curve crosses the x-axes at (-4, 0)

We know that the curve crosses the y-axis at (0, -3)

Notice that when our curve crosses the x-axis at (-4, 0), it goes from above the axis to below the axis.

How we know this?

Remember that "crossing" the x-axis means that the sign of f(x) changes.

At the value x = -2.5 (which is larger than x = -4) the function is negative.

f(-2.5) = -12

and:

f(-4) = 0

So we can see that after f(x) crosses the x-axis at x = -4, the function is negative.

This means that before that point, the function must be positive.

So for values of x smaller than -4, the function should be larger than zero.

f(x) > 0 if x < -4

From this, we can conclude that in the range (-∞, –4), the function is above the x-axis.

Then we would write this as:

f(x) > 0 over the interval (–∞, –4)

The correct option is the last option.

User Aditya Mittal
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3.1k points