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Question: How do you find the vertical asymptotes and holes of: \(f(x) = \dfrac{{x + 2}}{{{x^2} + 3x - 4}}\)...

How do you find the vertical asymptotes and holes of:
f(x)=x+2x2+3x−4f(x) = \dfrac{{x + 2}}{{{x^2} + 3x - 4}} ?

Explanation

Solution

For finding the vertical asymptotes we have to use the concept that Vertical asymptotes occur at the values where a rational function contains a zero denominator.For finding the horizontal asymptotes we have to use the concept that it occur when the numerator of a rational function has degree less than or equal to the degree of the denominator.Holes occur only when we have duplicate factors on the numerator and denominator. If there is no duplication function has no holes.For finding the slant asymptotes we have to use the concept that it occurs when the degree of the denominator of a rational function is one less than the degree of the numerator.

Complete step by step answer:
We have, f(x)=x+2x2+3x−4f(x) = \dfrac{{x + 2}}{{{x^2} + 3x - 4}}.We have to find vertical, horizontal and slant asymptotes,
Vertical asymptotes: Vertical asymptotes occur at the values where a rational function contains a zero denominator.Therefore, solve x2+3x−4=0{x^2} + 3x - 4 = 0 ,
x2+3x−4=0 ⇒(x+4)(x−1)=0 ⇒x=1,x=−4{x^2} + 3x - 4 = 0 \\\ \Rightarrow (x + 4)(x - 1) = 0 \\\ \Rightarrow x = 1,x = - 4
We have the required vertical asymptotes.

Horizontal Asymptotes: Horizontal Asymptotes occur when the numerator of a rational function has degree less than or equal to the degree of the denominator.
lim⁡x→±∞f(x)→c\mathop {\lim }\limits_{x \to \pm \infty } f(x) \to c
divide terms on numerator and denominator by x2{x^2},
we will get ,
f(x)=x+2x2+3x−4 ⇒f(x)=xx2+2x2x2x2+3xx2−4x2 ⇒f(x)=1x+2x21+3x−4x2 f(x) = \dfrac{{x + 2}}{{{x^2} + 3x - 4}} \\\ \Rightarrow f(x) = \dfrac{{\dfrac{x}{{{x^2}}} + \dfrac{2}{{{x^2}}}}}{{\dfrac{{{x^2}}}{{{x^2}}} + \dfrac{{3x}}{{{x^2}}} - \dfrac{4}{{{x^2}}}}} \\\ \Rightarrow f(x)= \dfrac{{\dfrac{1}{x} + \dfrac{2}{{{x^2}}}}}{{1 + \dfrac{3}{x} - \dfrac{4}{{{x^2}}}}} \\\
As x→±∞x \to \pm \infty ,
f(x)=0+01+0−0=0 ⇒y=0f(x) = \dfrac{{0 + 0}}{{1 + 0 - 0}} = 0 \\\ \Rightarrow y = 0
We have the required horizontal asymptote. There are no holes because there is no duplicate factor present.

Note: Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. An asymptote could be a line that the graph of a function approaches as either ‘x’ or ‘y’ attend positive or negative infinity. There are three forms of asymptotes: vertical, horizontal and oblique.