-1

First part is original equation, second is after it was divided using polynomial division to find equation for oblique asymptote.I don't understand the 'x-1' part being the equation of the asymptote here,and the rest can be ignored because its basically 0, but for it to be 0 you set x to be nearing infinity, so then doesnt the 'x -1' at the beginning basically = infinity. And then your equation is y = infinity

$$y= \frac{(x^2)}{x+1} , y= x - 1 + \frac{(1)}{x+1}$$

horizontal asymptotes are straight so y is always one value. If I were to plot the graph for the first equation some points on the oblique asymptote would be around like (1,-1/2) so what does the line even mean/show then.I saw someone say to think of

$y = x - 1 + \frac{(1)}{x+1}$

as a function and

$x+1$

as another then compare the two and so on but I dont get why I would do that.

amnna
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  • The function y is a hyperbola that admits 2 branches separated by a vertical asymptote at x=-1. The function is negative for x<-1 and positive for x>-1. The negative branch admits the line y = x-1 as an asymptote: it has a maximum located at x=-2 for which y=-4. The positive branch is asymptotic to the same line y = x-1 and has a minimum at x=0 for which y=0. You may calculate the derivative to find the extrema. – Jean-Luc Boulnois Mar 17 '24 at 01:15
  • An oblique line is an asymptote of a function $f(x)$ if the graph of $y=f(x)$ gets arbitrarily close to the oblique line when you are far from the origin. If the equation of the oblique line is $y=g(x)$, then $|f(x) - g(x)|$ is a reasonable way to measure how close the graph of $f$ is to the line. Now make $x$ very large and see what happens to $|f(x) - g(x)|$. – David K Mar 17 '24 at 01:55

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