Sketch the graph of $y = \frac{4x^2 + 1}{x^2 - 1}$ I need help sketching the graph of $y = \frac{4x^2 + 1}{x^2 - 1}$.  
I see that the domain is all real numbers except $1$ and $-1$ as $x^2 - 1 = (x + 1)(x - 1)$.  I can also determine that between $-1$ and $1$, the graph lies below the x-axis.
What is the next step?  In previous examples I have determined the behavior near x-intercepts.  
 A: You can simplify right away with 
$$
y = \frac{4x^2 + 1}{x^2 - 1} = 4+ \frac{5}{x^2 - 1} =4+ \frac{5}{(x - 1)(x+1)}
$$
Now when $x\to\infty$ or $x\to -\infty$, adding or subtracting 1 doesn't really matter hence that term goes to zero. When $x$ is quite large, say 1000, the second term is very small but positive hence it should approach to 4 from above (same holds for negative large values). 
The remaining part to be done is when $x$ approaches to $-1$ and $1$ from both sides. For the values $x<-1$ and $x>1$ you can show that the second term is positive and negative for $-1<x<1$. Therefore the limit jumps from $-\infty$ to $\infty$ at each vertical asymptote. 
Here is the whole thing.
A: There can be no $x$-intercepts since the numerator is never $0$.  There are vertical asymptotes at $\pm1$ since the denominator there is $0$ and the numerator is not.  There is a horizontal asymptote at $4$, in both directions, because when $x$ is large in absolute value, the lower-degree terms are negligible by comparison to $x^2$.  And the curve never touches the horizontal asymptote because if the fraction equals $4$, then $4x^2 + 1 = 4(x^2-1)$ and that implies $1=-4$ by canceling the $4x^2$ from both sides.
