Compare the graphs of the inverse variations. Please provide at least 3 comparisons
y= (-0.2/x) and y= (-0.3/x)
-0.05, -0.075 , -0.1, -0.15 ?
To compare the graphs of the inverse variations y = (-0.2/x) and y = (-0.3/x), we can examine their key characteristics.
1. Rate of change:
The rate of change in both equations is negative, indicating an inverse variation. However, the rate of change in the second equation, y = (-0.3/x), is greater than in the first equation, y = (-0.2/x). This means that the second equation decreases at a faster rate than the first equation as x increases.
2. Asymptotes:
Both equations have vertical asymptotes when x = 0, since division by zero is undefined. Thus, the graphs approach the y-axis but never actually touch it.
3. y-intercepts:
To find the y-intercepts, we can set x = 0 and solve for y. For the first equation, y = (-0.2/0) is undefined since division by zero is not possible. Similarly, for the second equation, y = (-0.3/0) is also undefined. This implies that neither equation intersects the y-axis.
Given the values -0.05, -0.075, -0.1, and -0.15, it seems that you want to compare the value of y at these specific x-values. Let's calculate them for each equation:
For y = (-0.2/x):
- x = -0.05
y = (-0.2/(-0.05)) = 4
- x = -0.075
y = (-0.2/(-0.075)) = 2.6667
- x = -0.1
y = (-0.2/(-0.1)) = 2
- x = -0.15
y = (-0.2/(-0.15)) = 1.3333
For y = (-0.3/x):
- x = -0.05
y = (-0.3/(-0.05)) = 6
- x = -0.075
y = (-0.3/(-0.075)) = 4
- x = -0.1
y = (-0.3/(-0.1)) = 3
- x = -0.15
y = (-0.3/(-0.15)) = 2
Comparing the values of y at these x-values, we can observe that for both equations, as the absolute value of x decreases, the value of y increases. However, the second equation, y = (-0.3/x), yields higher values for y compared to the first equation, y = (-0.2/x), at each x-value.
In summary, the main comparisons between the two inverse variation equations are: the rate of change, the asymptotes, and the values of y at specific x-values.
To compare the graphs of the inverse variations y = (-0.2/x) and y = (-0.3/x), we can analyze their behavior for different x-values.
1. When x = -0.05:
For y = (-0.2/x), substituting x = -0.05 gives y = (-0.2/(-0.05)) = 4.
For y = (-0.3/x), substituting x = -0.05 gives y = (-0.3/(-0.05)) = 6.
Thus, for x = -0.05, the second inverse variation y = (-0.3/x) has a greater value than the first inverse variation y = (-0.2/x).
2. When x = -0.075:
For y = (-0.2/x), substituting x = -0.075 gives y = (-0.2/(-0.075)) ≈ 2.67.
For y = (-0.3/x), substituting x = -0.075 gives y = (-0.3/(-0.075)) ≈ 4.
Again, the second inverse variation y = (-0.3/x) has a greater value than the first inverse variation y = (-0.2/x) for x = -0.075.
3. When x = -0.1:
For y = (-0.2/x), substituting x = -0.1 gives y = (-0.2/(-0.1)) = 2.
For y = (-0.3/x), substituting x = -0.1 gives y = (-0.3/(-0.1)) = 3.
Here, the second inverse variation y = (-0.3/x) still has a greater value than the first inverse variation y = (-0.2/x) for x = -0.1.
In summary, the second inverse variation y = (-0.3/x) consistently has greater values than the first inverse variation y = (-0.2/x) for multiple x-values in the given range (-0.05, -0.075, -0.1).