Table of Contents

Last modified on August 28th, 2023

Vertical asymptotes, or VA, are dashed vertical lines on a graph corresponding to the zeroes of a function y = f(x) denominator. Thus, the curve approaches but never crosses the vertical asymptote, as that would imply division by zero.

We get the VA of the function as x = c when x approaches a constant value c going from left to right, and the graph tends to infinity or –infinity, as shown below. Note that c is not part of the domain of f(x). Hence, we can say that the vertical asymptotes of a function are not part of its domain.

A function can have infinite vertical asymptotes, which can be negative and extend to infinity. Unlike horizontal asymptotes, the curve never crosses the vertical asymptote.

We can determine the VA of a function f(x) from its graph or equation.

**From a Graph**

When looking at the f(x) graph, if any parts appear to be vertical, they are probably vertical asymptotes. We can check by drawing a vertical line from the point that looks like a VA. The curve is not a VA if it touches any part of this line.

**From an Equation**

Without a graph, we can find the vertical asymptotes from the equation of the function itself. To do so, we need to find the values of x such that ${ \lim _{x\rightarrow c}f\left( x\right) \rightarrow \infty}$ or ${\lim _{x\rightarrow c}f\left( x\right) \rightarrow -\infty}$. The values of c obtained from solving the above will give us all the vertical asymptotes.

While this rule gives us all the vertical asymptotes for most functions, there are a few more straightforward methods for some cases, which we will discuss below.

For rational functions of the form ${f\left( x\right) =\dfrac{P\left( x\right) }{Q\left( x\right) }}$, where P(x) and Q(x) are polynomials, we can calculate the VA using these rules:

- First, factorize the numerator and denominator, respectively, of the rational function.
- Next, simplify by canceling the common terms between the numerator and denominator. This step is important as these terms do not give a vertical asymptote.
- Finally, set the simplified denominator to zero and solve the remaining terms for x. The zeroes of the denominator are thus the vertical asymptotes of the function f(x).

For example, let us take the function ${f\left( x\right) =\dfrac{x^{2}+4x+4}{x^{2}+x-2}}$.

On factoring the numerator and denominator, we get

${f\left( x\right) =\dfrac{\left( x+2\right) \left( x+2\right) }{\left( x-1\right) \left( x+2\right) }}$

${\Rightarrow f\left( x\right) =\dfrac{\left( x+2\right) }{\left( x-1\right) }}$

Thus, (x – 1) remains the denominator after removing the common terms. Solving this for zero, we get x = -1 as the vertical asymptote for f(x).

The common factor (x + 2) in the above example is a **r**emovable discontinuity or a hole. It is a point where the graph is undefined, indicated by an open circle. For f(x), the point (0,-2) is the removable discontinuity.

Of the six basic trigonometric functions, sine and cosine do not have any VA. The other four, namely tan, sec, cosec, and cot, all have vertical asymptotes for the following values of x.

- y = tan x has vertical asymptotes at ${x=n\pi +\dfrac{\pi }{2}}$, where ‘n’ is an integer.
- y = sec x has vertical asymptotes at ${x=n\pi +\dfrac{\pi }{2}}$, where ‘n’ is an integer.
- y = cosec x has vertical asymptotes at x = nπ, where ‘n’ is an integer.
- y = cot x has vertical asymptotes at x = nπ, where ‘n’ is an integer.

The vertical asymptote of any logarithmic function is found by setting its argument to zero and solving for x. From this rule, the VA of a standard logarithmic function ${ f\left( x\right) =\log _{a}x}$ or f(x) = ln x is x = 0.

- Rectangular hyperbolic functions of the form ${ f\left( x\right) =\dfrac{a}{x+p}+q}$ have a vertical asymptote at x = -p.
- Polynomial functions of the form ${f\left( x\right) =a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots +a_{1}x^{1}+a_{0}}$ do not have any vertical asymptote as x does not give any real values at limits of infinity and –infinity.
- Exponential functions of the form ${ f\left( x\right) =a^{x}}$ or ${f\left( x\right) =ba^{cx}+k}$ are defined for all real values of x and so have no vertical asymptotes.

**Calculate the vertical asymptote of the function****${f_{\left[ x\right] }=\dfrac{x^{2}+2x-35}{x^{2}+25-10x}}$.**

Solution:

Factoring the numerator and denominator, we get

${f\left( x\right) =\dfrac{\left( x+7\right) \left( x-5\right) }{\left( x-5\right) ^{2}}=\dfrac{\left( x+7\right) }{\left( x-5\right) }}$

Thus, we have (x – 5) as the remaining factor in the denominator. On equating with zero, x = 5 is the vertical asymptote.

**Find all the vertical asymptotes of the function ****${f\left( x\right) =\dfrac{\left( x+1\right) \left( 2x+1\right) }{\left( x+3\right) \left( x-2\right) }}$.**

Solution:

Since this is factored already, we need to equate the factors of the denominator with zero and solve for x. So, the vertical asymptotes for (x + 3) = 0 and (x – 2) = 0 are x = -3 and x = 2, respectively.

**Give the vertical asymptote of the function ${f\left( x\right) =5\ln \left( x-3\right)}$.**

Solution:

To find the vertical asymptote, we have to make the argument zero and solve for x. Here, x = 3 when (x – 3) = 0. So the vertical asymptote is the line x = 3.

**Identify the vertical asymptote(s) of the function ****${f\left( x\right) =\dfrac{x^{2}-9}{x^{2}+x-12}}$.**

Solution:

The given function is in standard form, so we must factorize it first. On factorizing, we have

${f\left( x\right) =\dfrac{x^{2}-9}{x^{2}+x-12}=\dfrac{\left( x+3\right) \left( x-3\right) }{\left( x+4\right) \left( x-3\right) }}$

${\Rightarrow f\left( x\right) =\dfrac{\left( x+3\right) }{\left( x+4\right) }}$

Now we have the only remaining term in the denominator as (x + 4). Putting it as zero and solving for x, we get x = -4 as the vertical asymptote of the function f(x).

Last modified on August 28th, 2023