1) Determine the horizontal and vertical asymptotes in case they exist from the graph of the function.

$f(x) = \frac{|x|+1}{x-2}$

By looking at the function I can't deduce what kind of asymptotes it has.

I checked the graph of this function with a program, and it has the 2 asymptotes.

Another thing I didn't understand, this $|x|$, does it mean the function has 2 horizontal asymptotes ? , because as I approach x from both + infinity and - infinity I can check in the graph that Y gets close to $-1$ and $1$.

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    $\begingroup$ Note that the presence of the absolute value means we can define $f(x)$ piecewise as follows: For $x \geq 0$, we have $\frac{x + 1}{x - 2}$, and for $x < 0$, we have $\frac{-x + 1}{x - 2}$. $\endgroup$ – David A. Lee Jul 9 '17 at 17:17
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    $\begingroup$ Small nitpick: Your function $f(x)$ is not rational, because its numerator is not a polynomial. But as David points out in the comment above, it's piecewise rational, and you can apply your usual facts about limits of rational functions to the pieces. $\endgroup$ – Matthew Leingang Jul 9 '17 at 17:23
  • $\begingroup$ @David Lee thanks for the help, Clear up something, according to the rule to determine the H.A, If the degree of the numerator is equal to the degree of the denominador in order get the H.A, I can divide the leading coeficients, But because of the absolute value if I follow the rule I'll get a positive value for the HA and a negative value for the H.A, does it mean there isn't H.A ? $\endgroup$ – Goun2 Jul 9 '17 at 18:36

Since we cannot factor out a $x-2$ term in the numerator for either of the $x \geq 0$ or $x < 0$ cases (the numerator is only a polynomial of degree $1$, after all), we can conclude that $x=2$ must be the sole vertical asymptote.

Now, as we discussed, the function $f(x)$ is defined to be piecewise rational as follows:

$$f(x) = \begin{cases} \frac{x+1}{x-2}, \quad x \geq 0 \\ \frac{-x+1}{x-2}, \quad x < 0 \\ \end{cases}$$

The definition of horizontal asymptote is as follows: we say that the function $f(x)$ has a horizontal asymptote $y = c$ if

$$\lim_{x \rightarrow +\infty} f(x) = c \quad \text{or} \quad \lim_{x \rightarrow -\infty} f(x) = c$$

The or part is what you should pay attention to, as we need only satisfy one of these conditions for $y = c$ to be a horizontal asymptote.

From this definition, we consider the piecewise definitions of $f(x)$ in each respective to domain to find the horizontal asymptotes. In your comment you said you can divide by the leading coefficient, but I think you meant to say that you divide by the leading variable (in this case, $x$) in both the numerator and denominator. This part is straightforward, and at once you can see that

$$y = 1 \quad \text{and} \quad y = -1$$

are indeed horizontal asymptotes as $x \rightarrow +\infty$ and $x \rightarrow -\infty$, respectively.


As noted in the comments your function is not rational, but only piecewise rational.

Anyway , a rational function of the form $$f(x)=\frac{p^n(x)}{q^m(x)}$$ where $m$ and $n$ are the degrees of the two polynomials, we have :

1) If $x=a$ is a root of $q^m(x)$ and it is not a root of $p^n(x)$ than $x=a $ is a vertical asymptote.

2) if $n<m$ than $y=0$ is an horizontal asymptote

3) if $n=m$ than the funtion has an horizontal asymptote $y=k\ne0$

4) if $n=m+1$ the function has an oblique asymptote.

  • $\begingroup$ I can't see my function being of that form your showed. $\endgroup$ – Goun2 Jul 9 '17 at 19:08

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