Question

How would you find the equations of both the horizontal and vertical asymptotes for the following functions:
(i) $y = \dfrac{x}{{x - 3}}$
(ii) $y = \dfrac{{x + 4}}{{{x^2} - 1}}$
(iii) $y = \dfrac{{{x^2} - 2x + 1}}{{{x^2} - 3x - 4}}$

Answer 1

According to the question, for finding vertical asymptotes, we have to equate the denominator to 0 and solve for ‘x’. For finding the horizontal asymptote, we have to divide both the numerator and the denominator by the highest degree of x, and then solve for y.

Complete step by step solution:
(i) The function is:
$y = \dfrac{x}{{x - 3}}$
To get the vertical asymptote, we will make the denominator equal to 0, and we get:
$\Rightarrow x - 3 = 0$
When we solve for ‘x’, we get:
$\Rightarrow x = 3$
We got our asymptote as $x = 3$. This tells us that the value of ‘x’ cannot be zero and the asymptote is a vertical asymptote. This is because the value of the numerator is not zero.
When we calculate for horizontal asymptotes, we get:
$\mathop {\lim }\limits_{x \to \pm \infty } {\text{,y}} \to {\text{c}}\,{\text{(constant)}}$
The given function is:
$y = \dfrac{x}{{x - 3}}$
We will divide both the numerator and the denominator by ‘x’, and we get:
$= \dfrac{{\dfrac{x}{x}}}{{\dfrac{{x - 3}}{x}}}$
We can also write it as:
$= \dfrac{{\dfrac{x}{x}}}{{\dfrac{x}{x} - \dfrac{3}{x}}}$
When we simplify it, we get that:
$= \dfrac{1}{{1 - \dfrac{3}{x}}}$
When we apply the value of ‘x’ in y, we get:
When $x \to \pm \infty$, then $y \to \dfrac{1}{{1 - 0}}$
$\Rightarrow y = 1$
This is the horizontal asymptote.
Therefore, the vertical asymptote is $x = 3$and the horizontal asymptote is $y = 1$.
(ii) The function is:
$y = \dfrac{{x + 4}}{{{x^2} - 1}}$
To get the vertical asymptote, we will make the denominator equal to 0, and we get:
${x^2} - 1 = 0$
When we solve it, we get:
$\Rightarrow (x + 1)(x - 1) = 0$
$\Rightarrow x = \pm 1$
So, we get that the asymptotes are:
$x = 1\,and\,x = - 1$
Both are the vertical asymptotes.
When we calculate for horizontal asymptotes, we get:
$\mathop {\lim }\limits_{x \to \pm \infty } {\text{,y}} \to {\text{c}}\,{\text{(constant)}}$
The given function is:
$y = \dfrac{{x + 4}}{{{x^2} - 1}}$
We will divide both the numerator and the denominator by the highest power of ‘x’ which is ${x^2}$, and we get:
$= \dfrac{{\dfrac{{x + 4}}{{{x^2}}}}}{{\dfrac{{{x^2} - 1}}{{{x^2}}}}}$
We can also write it as:
$= \dfrac{{\dfrac{x}{{{x^2}}} + \dfrac{4}{{{x^2}}}}}{{\dfrac{{{x^2}}}{{{x^2}}} - \dfrac{1}{{{x^2}}}}}$
$= \dfrac{{\dfrac{1}{x} + \dfrac{4}{{{x^2}}}}}{{1 - \dfrac{1}{{{x^2}}}}}$
When $x \to \pm \infty$, then $y = \dfrac{{0 + 0}}{{1 - 0}}$
$\Rightarrow y = 0$
This is the horizontal asymptote.
Therefore, the vertical asymptotes are $x = 1\,and\,x = - 1$. The horizontal asymptote is $y = 0$.
(iii) The given function is:
$y = \dfrac{{{x^2} - 2x + 1}}{{{x^2} - 3x - 4}}$
We will first factorize both the numerator and the denominator, and we get:
$y = \dfrac{{(x - 1)(x - 1)}}{{(x - 4)(x + 1)}}$
To get the vertical asymptote, we will make the denominator equal to 0, and we get:
$\Rightarrow (x - 4)(x + 1) = 0$
After solving, we get that:
$x = 4\,and\,x = - 1$
So, the vertical asymptotes are $x = 4\,and\,x = - 1$
When we calculate for horizontal asymptotes, we get:
$\mathop {\lim }\limits_{x \to \pm \infty } {\text{,y}} \to {\text{c}}\,{\text{(constant)}}$
The given function is:
$y = \dfrac{{{x^2} - 2x + 1}}{{{x^2} - 3x - 4}}$
We know that the highest degree of both the numerator and the denominator is 2. When the degree of the numerator and the denominator are the same, then the horizontal asymptote will be calculated on the basis of the ratio between the coefficients of the highest degree in both the numerator and the denominator. So, we get the horizontal asymptotes as:
$\dfrac{{1 \times {x^2}}}{{1 \times {x^2}}} = \dfrac{1}{1} = 1$
So, the horizontal asymptotes is $y = 1$.
Therefore, the vertical asymptotes are $x = 4$ and $x = - 1$. The horizontal asymptote is $y = 1$.

Note:
In (iii) when we were factorizing the given function, we got $y = \dfrac{{(x - 1)(x - 1)}}{{(x - 4)(x + 1)}}$. We can see that nothing is getting cancelled from the numerator and the denominator. This means that the function is not having any holes.