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5.6 Rational functions  (Page 4/16)

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Vertical asymptotes

The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.

Given a rational function, identify any vertical asymptotes of its graph.

  1. Factor the numerator and denominator.
  2. Note any restrictions in the domain of the function.
  3. Reduce the expression by canceling common factors in the numerator and the denominator.
  4. Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.
  5. Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities, or “holes.”

Identifying vertical asymptotes

Find the vertical asymptotes of the graph of   k ( x ) = 5 + 2 x 2 2 − x − x 2 .

First, factor the numerator and denominator.

k ( x ) = 5 + 2 x 2 2 − x − x 2 = 5 + 2 x 2 ( 2 + x ) ( 1 − x )

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

( 2 + x ) ( 1 − x ) = 0 x = −2 , 1

Neither   x = –2   nor   x = 1   are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in [link] confirms the location of the two vertical asymptotes.

Graph of k(x)=(5+2x)^2/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.
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Removable discontinuities

Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity    .

For example, the function   f ( x ) = x 2 − 1 x 2 − 2 x − 3   may be re-written by factoring the numerator and the denominator.

f ( x ) = ( x + 1 ) ( x − 1 ) ( x + 1 ) ( x − 3 )

Notice that   x + 1   is a common factor to the numerator and the denominator. The zero of this factor,   x = −1 ,   is the location of the removable discontinuity. Notice also that   x – 3   is not a factor in both the numerator and denominator. The zero of this factor,   x = 3 ,   is the vertical asymptote. See [link] . [Note that removable discontinuities may not be visible when we use a graphing calculator, depending upon the window selected.]

Graph of f(x)=(x^2-1)/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1.

Removable discontinuities of rational functions

A removable discontinuity    occurs in the graph of a rational function at   x = a   if   a   is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.

Identifying vertical asymptotes and removable discontinuities for a graph

Find the vertical asymptotes and removable discontinuities of the graph of   k ( x ) = x − 2 x 2 − 4 .

Factor the numerator and the denominator.

k ( x ) = x − 2 ( x − 2 ) ( x + 2 )

Notice that there is a common factor in the numerator and the denominator,   x – 2.   The zero for this factor is   x = 2.   This is the location of the removable discontinuity.

Notice that there is a factor in the denominator that is not in the numerator,   x + 2.   The zero for this factor is   x = −2.   The vertical asymptote is   x = −2.   See [link] .

Graph of k(x)=(x-2)/(x-2)(x+2) with its vertical asymptote at x=-2 and a removable discontinuity at x=2.

The graph of this function will have the vertical asymptote at   x = −2 ,   but at   x = 2   the graph will have a hole.

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Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
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Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
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8m/s²
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What is Thermodynamics
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Another formula for Acceleration
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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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