How to find out whether a function has vertical asymptotes
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 ?
$\endgroup$2 Answers
$\begingroup$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 $$x<0$$
$\endgroup$ 5 $\begingroup$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 $x \geq 0$
$$f(x)=\frac{16(x-4)}{x-4}, x\neq 4$$
For the second function, $x <0$
$$f(x)=\frac{16x-64}{-x-4},x \neq -4$$
