Posted by Meron on Thursday, November 10, 2011 at 8:00am.
If you meant
(x^3+1)/(x^2-1)
then if x = -1, your expression becomes 0/0 which is "indeterminate".
notice that (x^3+1)/(x^2-1)
= (x+1)(x^2 + x + 1)/((x+1)(x-1))
= (x^2 + x + 1)/(x-1)
I don't know if you know anything about limits, but
the limit of the above , as x --> - 1
= (1 - 1 + 1)/(-2) = -1/2
There are actually no zeros for this graph,
as
http://www.wolframalpha.com/input/?i=%28x%5E3%2B1%29%2F%28x%5E2-1%29
will show you.
the expression actually reduces to
y = (x^2 + x + 1)/(x-1) , which has a vertical asymptote at x=1
the original function
y = (x^3 + 1)(/(x^2-1) looks exactly the same except it would have a "hole" or missing point at (-1, -1/2)
Thank u so much!
Related Questions
math,correction, bobpursly - Since when is 5^7 =8125? if a number is a multple ...
Algebra - Hi this question was already asked but there is still no correct ...
math,correction - can someone correct this for me parts A-E and if i am wrong ...
Math Inequalities - -9/8x greater than or equal to -1/16 can someone please ...
math,correction,plz - Can someone who has a few minutes could you please correct...
math help please - if the product of two whole numbers is zero, then one of the ...
Math - ok so i'm kinda sort of confused... I had this for a equation X^2 + ...
Math - I need someone to please check my answer for me... add. -1.6 + (-2.3)= ...
maths - how do i solve this equation? there is another answer other than 0, but ...
Economics - The market demand in a Bertrand duopoly is P = 15 - 4Q, and the ...
For Further Reading
