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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.

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Sorry for my last comment. You've already factored the denominator. Locate the vertical asymptotes next. Make sure when you're doing this that you locate zeroes of the denominator that aren't zeroes of the numerator (this won't happen for your function) –  David Mitra Nov 15 '11 at 20:12
    
With the zeros (and their multiplicities), the vertical asymptotes (and their multiplicities), the slant/horizontal asymptote, and a few other points (from a table...) in hand, you should be able to "connect the dots". –  The Chaz 2.0 Nov 15 '11 at 20:28

2 Answers 2

up vote 3 down vote accepted

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.

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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.

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