
How do you find any asymptotes of \[f\left( x \right) = \dfrac{x}{{x - 5}}\]?
Answer
514.8k+ views
Hint: Here in this question, we have to find the asymptotes of the given rational function. To find the vertical asymptote of a rational function, set the denominator equal to zero and solve for x and the horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator. Degree of the numerator is less than the degree of the denominator: horizontal asymptote at y = 0.
Complete step by step answer:
An asymptote of the curve \[y = f(x)\] or in the implicit form: \[f(x,y) = 0\] is a straight line such that the distance between the curve and the straight line lends to zero when the points on the curve approach infinity.
There are three types of asymptotes namely:
Vertical Asymptotes - When \[x\] approaches some constant value \[c\] from left or right, the curve moves towards infinity (i.e., \[\infty \]) , or -infinity (i.e., \[ - \infty \]) and this is called Vertical Asymptote.
Horizontal Asymptotes - When \[x\] moves to infinity or -infinity, the curve approaches some constant value \[b\], and is called a Horizontal Asymptote.
Oblique Asymptotes - When x moves towards infinity (i.e., \[\infty \]), or -infinity (i.e., \[ - \infty \]), the curve moves towards a line \[y = mx + b\], called Oblique Asymptote.
Consider the given question
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{x}{{x - 5}}\]
The denominator of \[f\left( x \right)\] cannot be zero as this would make \[f\left( x \right)\] undefined. Equating the denominator to zero 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.
\[ \Rightarrow \,\,x - 5 = 0\]
On solve this equation for \[x\], we get
\[ \Rightarrow \,\,x = 5\] is the vertical asymptote
horizontal asymptotes occur as \[\mathop {\lim }\limits_{x \to \pm \infty } ,\,f(x) \to c\]
where c is a constant
consider the function
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{x}{{x - 5}}\]
Divide both numerator and denominator by \[x\], then
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{{\dfrac{x}{x}}}{{\dfrac{x}{x} - \dfrac{5}{x}}}\]
On simplification, we get
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{1}{{1 - \dfrac{5}{x}}}\]
Where, as \[x \to \pm \infty \] then
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{1}{{1 - \dfrac{5}{x}}}\]
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{1}{{1 - 0}}\]
On simplification, we get
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{1}{1}\]
\[ \Rightarrow \,\,f\left( x \right) = 1\]
or
\[ \Rightarrow \,\,y = 1\]
Which Is the vertical asymptote
Hence, the asymptotes of given functions \[f\left( x \right) = \dfrac{x}{{x - 5}}\] is \[x = 1\] and \[y = 1\].
Note: When we are finding the asymptotes of a function, we will apply the concept of the limit. When we apply the limit to the function, we obtain the real valued function. We have to know the different forms of asymptote and then by obtaining the final result, we obtain the type of asymptote.
Complete step by step answer:
An asymptote of the curve \[y = f(x)\] or in the implicit form: \[f(x,y) = 0\] is a straight line such that the distance between the curve and the straight line lends to zero when the points on the curve approach infinity.
There are three types of asymptotes namely:
Vertical Asymptotes - When \[x\] approaches some constant value \[c\] from left or right, the curve moves towards infinity (i.e., \[\infty \]) , or -infinity (i.e., \[ - \infty \]) and this is called Vertical Asymptote.
Horizontal Asymptotes - When \[x\] moves to infinity or -infinity, the curve approaches some constant value \[b\], and is called a Horizontal Asymptote.
Oblique Asymptotes - When x moves towards infinity (i.e., \[\infty \]), or -infinity (i.e., \[ - \infty \]), the curve moves towards a line \[y = mx + b\], called Oblique Asymptote.
Consider the given question
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{x}{{x - 5}}\]
The denominator of \[f\left( x \right)\] cannot be zero as this would make \[f\left( x \right)\] undefined. Equating the denominator to zero 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.
\[ \Rightarrow \,\,x - 5 = 0\]
On solve this equation for \[x\], we get
\[ \Rightarrow \,\,x = 5\] is the vertical asymptote
horizontal asymptotes occur as \[\mathop {\lim }\limits_{x \to \pm \infty } ,\,f(x) \to c\]
where c is a constant
consider the function
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{x}{{x - 5}}\]
Divide both numerator and denominator by \[x\], then
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{{\dfrac{x}{x}}}{{\dfrac{x}{x} - \dfrac{5}{x}}}\]
On simplification, we get
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{1}{{1 - \dfrac{5}{x}}}\]
Where, as \[x \to \pm \infty \] then
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{1}{{1 - \dfrac{5}{x}}}\]
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{1}{{1 - 0}}\]
On simplification, we get
\[ \Rightarrow \,\,f\left( x \right) = \dfrac{1}{1}\]
\[ \Rightarrow \,\,f\left( x \right) = 1\]
or
\[ \Rightarrow \,\,y = 1\]
Which Is the vertical asymptote
Hence, the asymptotes of given functions \[f\left( x \right) = \dfrac{x}{{x - 5}}\] is \[x = 1\] and \[y = 1\].
Note: When we are finding the asymptotes of a function, we will apply the concept of the limit. When we apply the limit to the function, we obtain the real valued function. We have to know the different forms of asymptote and then by obtaining the final result, we obtain the type of asymptote.
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