Sketch the region in the first quadrant enclosed by y=8/x,y=4x and y=1/4x
For y=4x, draw a straight line passing through (0,0) and (1,4).
For y=(1/4)x, draw a straight line passing through (0,0) and(4,1).
y=8/x is a reciprocal function. For just the first quadrant, sketch a smooth curve passing through (1,8), (2, 4), (4,2), and (8,1), with asymptotes at x=0 and y=0.
Shade in the region enclosed by the three graphs.
To sketch the region in the first quadrant enclosed by the equations y = 8/x, y = 4x, and y = 1/4x, you can follow these steps:
1. Plot the x-intercepts:
Since y = 8/x, when y = 0, x = 0. Therefore, there are no x-intercepts for this equation.
For y = 4x, when y = 0, x = 0. So there is an x-intercept at (0, 0).
For y = 1/4x, when y = 0, x = 0. So there is an x-intercept at (0, 0).
2. Determine the points where the different equations intersect each other.
To find the points of intersection, set the equations equal to each other and solve for x and y.
=> 8/x = 4x
=> 8 = 4x^2
=> x^2 = 2
=> x = ±√2
By substituting x into each equation, we can find the corresponding y-values:
For y = 8/x, when x = √2, y = 8/√2 = 4√2.
For y = 8/x, when x = -√2, y = 8/-√2 = -4√2.
For y = 4x and y = 1/4x equations, we can substitute x = √2 and x = -√2 and solve for y.
3. Plot the points and sketch the region bounded by the curves.
The determined points are (0, 0), (√2, 4√2), (-√2, -4√2), and (0, 8).
Since we are only sketching the region in the first quadrant, we need to exclude the points (-√2, -4√2) and (0, 0) from our region.
Therefore, the region enclosed by the curves y = 8/x, y = 4x, and y = 1/4x in the first quadrant is a triangular region with three vertices at (0, 0), (√2, 4√2), and (0, 8).