i have the function $f(x) = \frac {(x+1)(x-1)^2}{x^2}$ and i want to sketch it so i found the following: Vertical Asymptote: is x=0 X-intercepts: x=1, x=-1, x=1 Slant Asymptote: $y = x-1 $
then i wanted to see if the graph intersected the Slant asymptote so i set the function equal to the slant aymptote and solved for x so: $x-1 = \frac {(x+1)(x-1)^2}{x^2}$ and i got $ (x^2 -2) $ and $(x-1)$ and factors so i solved for x and got x=1 and x= 2 and x=-2
but how do i graph there points ? what do these points mean ?
Your oblique asymptote equation is correct, but your work is wrong. You should get $x=1$ as your $x$ coordinate for the point of intersection.
To find the $y$ coordinate, simply plug in $x=1$ to either equation and you'll see that the point of intersection is $(1,0)$. Graphing this should be self-explanatory.
With horizontal and slant asymptotes, the function itself can cross these equations, but as its domain approached $-\infty$ and $\infty$, its graph approaches the equation of the asymptote. The fact that there is an intersection point simply means your particular equation crosses its asymptote, usually indicating a higher degree equation.
I hope this explains the "why" part of your question.