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Last modified on August 25th, 2023
An oblique or slant asymptote is a dashed line on a graph, describing the end behavior of a function approaching a diagonal line where the slope is neither zero nor undefined.
Thus, when either ${\lim _{x\rightarrow \infty }f\left( x\right)}$ or ${\lim _{x\rightarrow -\infty }f\left( x\right)}$ give the equation of a line mx + b, where m ≠ 0, then we say that the function f(x) has an oblique asymptote equal to mx + b, as shown below.
A function may have a horizontal or an oblique asymptote; it cannot have both. Like horizontal asymptotes, oblique asymptotes can cross the function. They can be negative and extend to infinity. Also, a function may have a maximum of two oblique asymptotes but can have infinitely many vertical asymptotes.
In general, for a function f(x), the oblique asymptote is a line l such that ${\lim _{x\rightarrow \pm \infty }\left( f\left( x\right) -l\left( x\right) \right) =0}$, or ${\lim _{x\rightarrow -\infty }\left( f\left( x\right) -l\left( x\right) \right) =0}$. To do this, we take l = mx + b and find m and b, which are the slope and the y-intercept, respectively, using these steps:
Thus, we get the equation of line l, which is the oblique asymptote of f(x). However, there are a few more direct methods to find the oblique asymptotes of various functions, which we will discuss below.
Rational functions of the form ${f\left( x\right) =\dfrac{P\left( x\right) }{Q\left( x\right) }}$, where Q(x) ≠ 0, f(x) will have an oblique asymptote only when the degree of P(x) is greater than Q(x) by one.
We can get the oblique asymptote using either of two rules:
For instance, the function ${f\left( x\right) =\dfrac{x^{2}-25}{x-5}}$ can be simplified to give f(x) = (x + 5). So, it has an oblique asymptote of y = x + 5.
Let us look at the function ${f\left( x\right) =\dfrac{2x^{2}+3x-1}{2x-1}}$. Since the denominator is not of the form (x + a), we cannot perform synthetic division. So, we must use long division to find the quotient.
We obtain (x + 2) as the quotient of the long division. The remainder is ignored. Thus, y = x +2 is the oblique asymptote of ${f\left( x\right) =\dfrac{2x^{2}+3x-1}{2x-1}}$.
Now let us take the function ${f\left( x\right) =\dfrac{2x^{2}+x-2}{x+2}}$. Here, the denominator corresponds to (x + a), so we may use synthetic division to find the quotient.
This method gives (2x + 3) as the quotient. Again, the remainder is ignored. Thus, we have y = 2x + 3 as the oblique asymptote of ${f\left( x\right) =\dfrac{2x^{2}+x-2}{x+2}}$.
Once we get the equation of the oblique asymptote, the last step is graphing it. To do this, we follow these steps:
Here, the function ${f\left( x\right) =\dfrac{2x^{2}-4}{x+3}}$ has an oblique asymptote y = 2x – 6. The y and x-intercepts are (0, -6) and (3, 0). The graph thus looks like this:
Give the oblique asymptote of the function ${f\left( x\right) =\dfrac{x^{2}+1}{x-1}}$ .
We get the oblique asymptote using long division.
The quotient is (x + 1). Thus, the oblique asymptote is y = x +1.
Calculate the oblique asymptotes of the hyperbola ${\dfrac{x^{2}}{25}-\dfrac{y^{2}}{9}=1}$ .
In the given hyperbola, a2 = 25 and b2 = 9. So, a = 5 and b = 3, respectively. The oblique asymptotes are thus y = (3/5)x and y = -(3/5)x.
Find the oblique asymptote of ${f\left( x\right) =\dfrac{x^{2}+4x+4}{x-1}}$ .
We will perform a long division to find the asymptote.
We get the quotient as (x + 5). The oblique asymptote is thus y = x + 5.
Last modified on August 25th, 2023