Theory :

Asymptote : An asymptote is a line that the graph of a function approaches but never reaches. \ \

There are two main types of asymptotes: 1) Horizontal 2) Vertical

1) Horizontal Asymptotes : \ \

To find a function\\'s horizontal asymptotes, there are 3 situations.

a.The degree of the numerator is higher than the degree of the denominator. In this case, then there are no horizontal asymptotes.

b.  The degree of the numerator is less than the degree of the denominator.In this case, then the horizontal asymptote is y=0.

c.  The degree of the numerator is the same as the degree of the denominator.

In this case, then the horizontal asymptote is y = ^a/_d.

where a is the coefficient in front of the highest degree in the numerator and d is the coefficient in front of the highest degree in the denominator. \ \

2) Vertical Asymptotes : \ \

1.  To find a vertical asymptote, set the denominator equal to 0 and solve for x \ \

Discontinuity : A discontinuity is a break in the graph. \ \

There are two main types of discontinuities.

1) Removable discontinuities 2) Non - Removable discontinuities \ \

1)Removable Discontinuities:  A discontinuity is a part of the graph that is undefined at a particular point, but there is no asymptote at that point. A removable discontinuity is when you can factor out a term in the numerator and factor out the same term in the denominator, thus canceling each other out. \ \

2)Non - Removable discontinuities : Make denominator equals to zero then solve x. \ \

Problem

Step 1 : \ \

Given function is  : (2x³ - 0.6x² - 8.6x - 6) / (3x³ - 0.6x² - 12.24x - 8.64) \ \

Numerator: 2x³ - 0.6x² - 8.6x - 6

Denominator: 3x³ - 0.6x² - 12.24x - 8.64

Identify Rational Zeros :

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation

a_nx ⁿ + a_n – 1x ^{n – 1} + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

The function Numerator is 2x ³- 0.6x ²- 8.6x - 6 = 0.

If p/q is a rational zero, then p is a factor of 6 and q is a factor of 1.

The possible values of p  are   ± 1, ± 2, and ± 3.

The possible values for q are   ± 1, ± 2

By the Rational Roots Theorem, the only possible rational roots are, p / q = ± 1,   ± 2,  ± 3 , ±1/2 , ± 3/2

Make a table for the synthetic division and test possible real zeros.

Make a table for the synthetic division and test possible real zeros.

Since, f(1) = 0, x = 1 is a zero. The depressed polynomial is  x ² - 1.6x - 6 = 0.

⇒ x = -1.2 and x = 2.5

Therefore, tthe roots of the function are x = -1, x = - 1.2, and x = 2.5

The function Denominator is 3x ³- 0.6x ²- 12.24x - 8.64 = 0.

Denominator: 3x³ - 0.6x² - 12.24x - 8.64

Substitute -1  in Denominator :

3(-1)³ - 0.6(-1)² - 12.24(-1) - 8.64

= - 3 - 0.6 + 12.24 - 8.64

= 0

Synthetic substitution p/q = -1

Make a table for the synthetic division and test possible real zeros.

Since, f(-1) = 0, x = -1 is a zero. The depressed polynomial is  3x ² - 3.6x - 8.64 = 0.

⇒ x = -1.2 and x = 2.4

Therefore, the roots of the function are x = -1, x = - 1.2, and x = 2.4 \ \

Step 2 : \ \

Given function is  : (2x³ - 0.6x² - 8.6x - 6) / (3x³ - 0.6x² - 12.24x - 8.64) \ \

Step 3 : \ \

To find a horizontal asymptote, \ \

The degree of the numerator is the same as the degree of the denominator. \ \

In this case, then the horizontal asymptote is 1/1 = 1. \ \

To find a vertical asymptote, \ \

set the denominator equal to 0 and solve for x \ \

x - 2.4 = 0 \ \

x = 2.4 \ \

To find Removable discontinuity, \ \

Canceled factors are ( x +1 )  and  (  x + 1.2 ) \ \

( x +1 )  = 0 and  (  x + 1.2 ) = 0 \ \

x  = -1  and x = -1.2 \ \

To find Non - Removable discontinuities, \ \

set the denominator equal to 0 and solve for x

x - 2.4 = 0

x = 2.4 \ \

Solution : \ \

Horizontal asymptote : 1 \ \

Hertical asymptote x = 2.4 \ \

Removable discontinuities  x  = -1  and x = -1.2 \ \

Non - Removable discontinuities x = 2.4