Pre-Calculus help please!!!?

Question

For what value(s) of x does the graph of the relation y= -2x^(3)-x^(2)+6x lie below the x-axis ??


please explain it to me, thanks in advance.

Ron W, thank you so much, Im working on it and once I get the answer(s) and check them out in the class, you will get your very deserved 10 points, very well explained, I really appreciate your help.=D

Answer 1

First, you need to find the x-values for which y = 0.

Factor the relation:

y = -2x³ - x² + 6x = -x(2x² + x - 6) = -x(2x - 3)(x + 2)

So y = 0 for x = 0, x = 3/2. or x = -2.

Plot these x-values on an x-axis or a number line. They divide the x-axis into intervals within which y always has the same sign.

So we have something like this:

--- -2 ----- 0 --- 3/2 ------> x

For each of the four regions shown (x < -2, -2 < x < 0, 0 < x < 3/2, x > 3/2), choose an x-value in the region and evaluate y at that value. If you get a negative number, then y is negative throughout that region (and hence the graph lies below the x-axis for this region); if you get a positive number, then y is positive throughout that region (and hence the graph lies above the x-axis for this region).

First, consider the region x < -2. Pick an x-value in that region, for example,
x = -3. Now, y(-3) = 27. So y > 0 for x < -2 (and so its graph lies above the x-axis
for x < -2).

Next, -2 < x < 0. Pick, say, x = -1. y(-1) = -5. So y < 0 for -2 < x < 0 (and so its graph lies below the x-axis for -2 < x < 0)

I will leave the other two regions to you.