Suggestions for graph analysis

In summary, when n is an odd number, the graph has two lines that are diagonal. When n is even, the graph only has one line.
  • #1
Boogieface
7
0

Homework Statement


fn(x) = x^-n is the given formula.
for parameter values n>0 on the domain x>0.
I am asked to substitute 4 different values of the parameter n, when n>1 and 0<n<1 and then explain how n changes the shape of the graph.

Homework Equations



The Attempt at a Solution


I have substituted 0.5, 0.95, 2 and 10 into n but the problem is trying an odd number would give me a different shape of graph unlike 2's and 10's. So, I don't know ideally what number is the best to explain their properties. My explanation for the graphs seem very simple because I could only see when n increases(n>1) the graph goes steeper and steeper. I have also found the vertical asymptote is undefined and the horizontal asymptote is indefinite.
What more could I add to explain the differences / similarities?
 
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  • #2
What difference do you see in the graph if n is odd? I don't think there are many differences for x>0 there.

Your graph analysis is ok, the graph goes steeper and steeper as n gets larger. But you could also mention what happens if n goes to 0. What happens to the graph if n is very close to 0?
 
  • #3
micromass said:
What difference do you see in the graph if n is odd? I don't think there are many differences for x>0 there.

Your graph analysis is ok, the graph goes steeper and steeper as n gets larger. But you could also mention what happens if n goes to 0. What happens to the graph if n is very close to 0?
the difference is just how it is drawn when n is odd, like if you put n = 3 the graph, two lines are drawn and they are diagonal ( I don't know if diagonal is the right word to describe ). If you put n = 2, you end up the graph with one line. that's what I was trying to say. Well, if n goes to 0, the I could only conclude y is equal to 1. if n is very close to 0, i guess the line will become a straighter and straighter line
Thank you for your suggestions
 

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