What are the vertical asymptotes of f(x) = 10/ x^2 - 1
1 answer:
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Answer:
We have the function:
f(x) = x^3 - 2*x^2
To sketch this, we need to graph some points, and then just draw a line that passes through the points.
The graph of this equation is shown below.
Now we can complete the question.
If the graph is below the x-axis in some interval, the function is negative in that interval
If the graph is above the x-axis in some interval, the function is positive in that interval.
If the graph goes up in a interval, then the function is increasing in that interval
If the graph goes down on an interval, then the function is decreasing in that interval.
Then:
1) f is------ on the intervals (−∞, 0) and (0, 2).
Here we can see that the graph is below the x-axis in those intervals, then here we have:
f is negative on the intervals (−∞, 0) and (0, 2).
2) f is------ on the interval (2,∞)
Here the answer is positive:
f is positive on the interval (2,∞)
3) fi is ------ on the interval (0, 4/3)
In the graph, you can see that the graph goes down in that interval, then the correct answer here is:
f is decreasing on the interval (0, 4/3)
4) f is------ on the intervals (−∞, 0) and (4/3, ∞).
In this case, we can see that the graph goes up in these intervals, then the correct answer here is:
f is increasing on the intervals (−∞, 0) and (4/3, ∞).
