The geometry of the derivative of y=(x^2-1)/(x^2-4) : sketching

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Jeane

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Without finding inflexions, sketch the graph of the following function. Indicate any stationary points and intercepts with the axes. y=(x^2-1)/(x^2-4)
y'=( v.du/dx-u.dv/dx)/v^2=(-6x)/ (x^2-4)^2 Vertical asymptotes at x^2-4=0 ie at x=-2 and x=2 Intercepts at x^2-1=0 ie at x=-1 and x=1 I don't know how to find the horizontal asymptotes but I do know it means finding what y equals. Drawing the curve is hard to imagine without knowing when it is concave up and concave down which you find from the second derivative. Thanks for your help
 
Jeane said:
I don't know how to find the horizontal asymptotes …
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Horizontal asymptotes occur at global behavior; that is, when x becomes very large in absolute value.

Can you determine the limit of y, as x approaches infinity? How about as x approaches negative infinity?

There are some simple rules to determine global behavior, for functions like this (i.e., a ratio of polynomials). They have to do with comparing the degree of the polynomials. In this case, the degree of each is the same. Have you seen the rule for that? :cool:
 
mmm4444bot said:
Horizontal asymptotes occur at global behavior; that is, when x becomes very large in absolute value.

Can you determine the limit of y, as x approaches infinity? How about as x approaches negative infinity?

There are some simple rules to determine global behavior, for functions like this (i.e., a ratio of polynomials). They have to do with comparing the degree of the polynomials. In this case, the degree of each is the same. Have you seen the rule for that? :cool:
Click to expand...
Not sure what you mean but have done some work on limits but didn't know it pertained to this situation. IN this case from what you have said y approaches two but that is not the answer given on the text.
 
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Jeane said:
… y approaches two …
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That's not correct. How did you calculate the limit of y as x approaches infinity? Please show your work, so that I can see how you got 2.

In the meanwhile, I will search online for a lesson on the rules.
 
Jeane said:
Not sure what you mean but have done some work on limits but didn't know it pertained to this situation. IN this case from what you have said y approaches two but that is not the answer given on the text.
Click to expand...
Sorry that is wrong it approaches 1/4? But that is still not right.
 
The limit of y as x approaches infinity is not 1/4.

It seems like you want to keep your work a secret; hence, I am not able to comment further.

Please read the forum guidelines.
 
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