what values of x does the graph of the relation y=-2x^3-x^2+6x lie below the xaxis?

Answer 1

First find the values for x that make this equation 0:

y = -2x^3 - x^2 + 6x
0 = x(-2x^2 - x + 6)
0 = x(2x - 3)(-x - 2)
x = 0, 3/2, -2

From here, we can see that between these points the graph is either above or below the x-axis. So let's plug in numbers to find out where the graph is.

Let's plug in something to the left of -2, like -1,000,000:

y = -2x^3 - x^2 + 6x
y = -2(-1,000,000)^3 + (-1,000,000)^2 + 6(-1,000,000)
y = a big positive number + a not-so-big negative number - a smaller negative number

The first term (-2x^3) controls where this curve will end up - as a positive number. so everything left of -2, or x<-2, is positive.

Moving right along, let's go from -2 to 0 - pick -1:

y = -2x^3 - x^2 + 6x
y = -2(-1)^3 - (-1)^2 + 6(-1)
y = 2 - 1 - 6
y = some negative number

So in this interval, -2
Next - between 0 and 3/2. 1 is an easy one to work with:

y = -2x^3 - x^2 + 6x
y = -2(1)^3 - (1)^2 + 6(1)
y = -2 - 1 + 6
y = some positive number

So, 0
Finally, look at everything right of 3/2, like 100:

y = -2x^3 - x^2 + 6x
y = -2(100)^3 - (100)^2 + 6(100)
y = some big negative number - a smaller negative number + a smaller positive number

The same rationale applies here as the first case. From here, we can see that anything right of 3/2, or x>3/2, is negative.

So your complete answer is -23/2 produces negative y-values and causes the curve to fall below the x-axis.