Slope Field
Last updated
Last updated
▶ Refer to Khan academy: Worked example: range of solution curve from slope field
Let's only try the point (2, 2)
, we will get y' = 0
, which means the line in Q1 would be horizontal.
Since 6
is the initial condition, so we make 6
as a boundary.
The slope at (0,6)
is a negative slope, which seems keeps negative until 4
.
So the Range would be (4, 6]
.
Try out a point, etc, the initial point is (0, 1)
.
TAKE DERIVATIVE of each equation, to get the SLOPE of each.
Plug in the x=0
value to get each slope, and eyeball it.
Solve:
Solve: Hint: Try a point or points in each quadrant, like Q1: (2,2), Q2: (-2,2), Q3: (-2,-2), Q4: (2,-2)
Solve:
Solve: