00:02
Okay, so we need to take each of the two parts of our inequality on the left side, the numerator and the denominator, and set them equal to zero to find our critical points.
00:12
So when we do that, we get x as equal to negative three, and we get x as equal to two.
00:17
And those two numbers make up our critical points in our interval here.
00:25
All right.
00:26
So when we do this, we have negative infinity to negative three, and i'll just use negative four for that one.
00:34
When i plug in negative 4 into my initial inequality, two times negative 4 is negative 8.
00:40
Negative 8 plus 6 is negative.
00:43
So i have a negative divided by negative minus 2 is a negative.
00:48
So that will come out to give me a net positive.
00:51
Right from negative 3 to 2, i will use 0.
00:54
So when i plug in 0 for the axis, i have 2 times 0 is 0 plus 6.
01:00
So i have a positive.
01:02
And then 0 minus 2 is the negative...