Question

How do you find vertical, horizontal and oblique asymptotes for $y = \dfrac{{2{x^2} + x + 2}}{{x + 1}}$?

Answer 1

In this question we have to find the asymptotes of the given function, first we will find the vertical asymptotes by using the fact that the vertical asymptotes will occur at those values of x for which the denominator is equal to zero so, equate denominator to zero, now for the horizontal asymptotes we will use the fact that , when the degree of the numerator is larger than the degree of the denominator, then the function will have no horizontal asymptote here compare the degrees of the numerator and denominator and for the oblique asymptote we must divide the given function , then we will get the oblique asymptotes.

Complete step by step solution:
Given function is $y = \dfrac{{2{x^2} + x + 2}}{{x + 1}}$,
Now we know that the denominator of $y$ cannot be zero as this would make $y$ undefined. Equating the denominator to zero i.e., $x + 1$ and solving gives the value that $x$ cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.
$\Rightarrow x + 1 = 0$,
Now subtract 1 from both sides of the equation we get,
$\Rightarrow x + 1 - 1 = 0 - 1$,
Now simplifying we get,$y = 0$
$\Rightarrow x = - 1$
So, the vertical asymptote will be $x = - 1$,
Now for the horizontal asymptote we know that when $x$goes to $\infty$or $- \infty$ then the function approaches to a constant, this can be written as,
$\Rightarrow \mathop {\lim }\limits_{x \to \pm \infty } y \to c$,
Now divide all terms on numerator/denominator by the highest power of $x$, that is ${x^2}$,
$\Rightarrow y = \dfrac{{\dfrac{{2{x^2}}}{{{x^2}}} + \dfrac{x}{{{x^2}}} + \dfrac{2}{{{x^2}}}}}{{\dfrac{x}{{{x^2}}} + \dfrac{1}{{{x^2}}}}}$,
Now simplifying we get,
$\Rightarrow y = \dfrac{{2 + \dfrac{x}{{{x^2}}} + \dfrac{2}{{{x^2}}}}}{{\dfrac{1}{x} + \dfrac{1}{{{x^2}}}}}$,
Now as $x$ goes to $\infty$or $- \infty$, then
$\Rightarrow y = \dfrac{{2 + 0 + 0}}{{0 + 0}}$,
Now simplifying we get,
$\Rightarrow y = \dfrac{2}{0} = \infty$,
This is undefined hence there are no horizontal asymptotes.
Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is the case here numerator-degree is 2 and denominator- degree 1,Hence there is an oblique asymptote.
Now using polynomial division we get,
$\Rightarrow \left. {x + 1} \right)2{x^2} + x + 2\left( {2x} \right. - 1$
$\underline {2{x^2} + 2x}$
$- x + 2$
$\underline { - x - 1}$
$3$
This can be written as,
$\Rightarrow y = 2x - 1 + \dfrac{3}{{x + 1}}$,
Now as $x$ goes to $\infty$or $- \infty$, then $y$ becomes,
$\Rightarrow y = 2x - 1 + 0$,
Now simplifying we get,
$\Rightarrow y = 2x - 1$,
So, the oblique asymptote will be $y = 2x - 1$.

**
$\therefore$ The asymptotes for the given function i.e., $y = \dfrac{{2{x^2} + x + 2}}{{x + 1}}$ will be
Vertical Asymptote will be $x = - 1$, there will be no horizontal asymptote and the oblique asymptote will be $y = 2x - 1$.**

Note:
An asymptote of a curve is the line formed by the movement of the curve and line moving continuously towards zero. In other words Asymptote is a line that a curve approaches as it moves towards infinity. Some facts on asymptotes are:
1. If the degree of the denominator is greater than the degree of the numerator, horizontal asymptote is at $y = 0$.
2. If the degree of the denominator is less than the degree of the numerator by one then we will get an oblique asymptote.
3. If the degree of the denominator is equal to the degree of the numerator, horizontal asymptote at the ratio of leading coefficients.