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I got the following question wrong in my exame

$\ \frac{16x-64}{|x|-4}$

My answer was, -16,16 for the horizontal asymptotes and 4,-4 for the vertical asymptotes.

For the horizontal asymptotes the answer is right, But there aren't any vertical asymptotes.

I thought if there wasn't a sum of squares in the denominator There would be vertical asymptotes.

So I proceeded this way, as I need to equal the denominator to 0 in order to find the vertical asymptotes.

$|x| - 4 = 0$

$x - 4 = 0$

$x = 4$


$-x - 4 = 0$

$-x = 4(-1)$

$x = -4$

There are 3 rules to find the HA's, but what are the rules to find the VA's ?

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note that you will have $$\frac{16x-64}{|x|-4}=\frac{16(x-4)}{x-4}$$ if $$x\geq0$$ or $$\frac{16(x-4)}{-x-4}$$ if $$x<0$$

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  • $\begingroup$ So, there are no vertical asymptotes because you canceled the denominator ? $\endgroup$ – Goun2 Jul 19 '17 at 19:30
  • $\begingroup$ for $$x<0$$ we have $$x=-4$$ as a vertical asymptote $\endgroup$ – Dr. Sonnhard Graubner Jul 19 '17 at 19:32
  • $\begingroup$ $x=+4$ is not a VA because the numerator also vanishes there, $f(x)$ has a removable discontinuity at $x=+4$ $\endgroup$ – WW1 Jul 19 '17 at 19:36
  • $\begingroup$ And how did you come to this conclusion, -4 is a VA ? you can cancel (x-4) with -(x-4), and the result would be 16/-1, There wouldn't be no function to equal to 0, sorry if i'm too dumb. $\endgroup$ – Goun2 Jul 19 '17 at 19:52
  • $\begingroup$ no $$-x-4=-(x+4)$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 19 '17 at 19:53
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We have the following function, $$f(x)=\frac{16x-64}{|x|-4}$$

The mathematical definition of absolute value of x is defined as

$$|x|\begin{cases}x& ,x \geq 0\\-x &, x<0\end{cases}$$

The absolute value of something will always the it positive. If it is negative, it will be negate to become positive.

Vertical asymptote of a rational function occurs when denominator is becoming zeroes. If a function like any polynomial $y=x^2+x+1$ has no vertical asymptote at all because the denominator can never be zeroes.

So for the above function we again have piece-wise function,

$$f(x)\begin{cases}\frac{16x-64}{x-4}& ,x \geq 0\\\frac{16x-64}{-x-4} &, x<0\end{cases}$$

For the domain of $x \geq 0$

$$f(x)=\frac{16x-64}{x-4}, x \neq 4$$

It can be further redefined such that,

$$f(x)=\frac{16(x-4)}{x-4}, x \neq 4$$

$$f(x)=16 ,x \neq 4$$

So, the first function has removable discontinuity at the point $x=4$. This is because $$\lim_{x \rightarrow 4^+}16=\lim_{x \rightarrow 4^-}16=\lim_{x \rightarrow 4}16$$

Removable discontinuity is defined such that

$$\lim_{x \rightarrow a^+}f(x)=\lim_{x \rightarrow a^-}=\lim_{x \rightarrow a}f(x)=L$$

although $x \neq a$.

However, if $x$ is defined on $a$ then there is no removable discontinuity. Notice that from the right side and the left side of a we are approaching same value.

For the second function with domain of $x<0$

$$f(x)=\frac{16x-64}{-x-4}, x\neq -4$$

Can it be further redefined? Obviously, we cannot cancel out the numerator and the denominator.

So, we know that it has vertical asymptote at some point when the denominator tends to $0$.Its denominator is ${-x-4}$

Taking the denominator and let it goes to zeroes

$$-x-4\rightarrow 0$$

$$-x\rightarrow4$$

$$x\rightarrow-4$$

So, the vertical asymptote occurs at $x=-4$ but wait is it inside the domain of the function? The answer is yes, because $-4<0$.

Let us use limit to see what is happening around it.

Right Hand Limit $$\lim_{x \rightarrow -4^+}\frac{16x-64}{-x-4}=- \infty$$

Left Hand Limit $$\lim_{x \rightarrow -4^-}\frac{16x-64}{-x-4}=+ \infty$$

$\lim_{x \rightarrow -4}f(x) $ does not exist in this case since $LHL \neq RHL$.

This type of discontinuity is called infinite discontinuity. As you can see it tends to $\pm \infty$ as it gets closer and closer to $-4$.

Some visual confirmation aid.

For the first function $x \geq 0$

$$f(x)=\frac{16(x-4)}{x-4}, x\neq 4$$

enter image description here

For the second function, $x <0$

$$f(x)=\frac{16x-64}{-x-4},x \neq -4$$

enter image description here

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