x3-1×2-9

Find where the expression (x-1)(x2+x+1)(x+3)(x-3) is undefined.

x=-3,x=3

Since (x-1)(x2+x+1)(x+3)(x-3)→-∞ as x→-3 from the left and (x-1)(x2+x+1)(x+3)(x-3)→∞ as x→-3 from the right, then x=-3 is a vertical asymptote.

x=-3

Since (x-1)(x2+x+1)(x+3)(x-3)→-∞ as x→3 from the left and (x-1)(x2+x+1)(x+3)(x-3)→∞ as x→3 from the right, then x=3 is a vertical asymptote.

x=3

List all of the vertical asymptotes:

x=-3,3

Consider the rational function R(x)=axnbxm where n is the degree of the numerator and m is the degree of the denominator.

1. If n<m, then the x-axis, y=0, is the horizontal asymptote.

2. If n=m, then the horizontal asymptote is the line y=ab.

3. If n>m, then there is no horizontal asymptote (there is an oblique asymptote).

Find n and m.

n=3

m=2

Since n>m, there is no horizontal asymptote.

No Horizontal Asymptotes

Simplify the expression.

Simplify the numerator.

Rewrite 1 as 13.

x3-13×2-9

Since both terms are perfect cubes, factor using the difference of cubes formula, a3-b3=(a-b)(a2+ab+b2) where a=x and b=1.

(x-1)(x2+x⋅1+12)x2-9

Simplify.

Multiply x by 1.

(x-1)(x2+x+12)x2-9

One to any power is one.

(x-1)(x2+x+1)x2-9

(x-1)(x2+x+1)x2-9

(x-1)(x2+x+1)x2-9

Simplify the denominator.

Rewrite 9 as 32.

(x-1)(x2+x+1)x2-32

Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=x and b=3.

(x-1)(x2+x+1)(x+3)(x-3)

(x-1)(x2+x+1)(x+3)(x-3)

(x-1)(x2+x+1)(x+3)(x-3)

Expand (x-1)(x2+x+1).

Apply the distributive property.

x(x2+x+1)-1(x2+x+1)(x+3)(x-3)

Apply the distributive property.

x(x2+x)+x⋅1-1(x2+x+1)(x+3)(x-3)

Apply the distributive property.

x⋅x2+x⋅x+x⋅1-1(x2+x+1)(x+3)(x-3)

Apply the distributive property.

x⋅x2+x⋅x+x⋅1-1(x2+x)-1⋅1(x+3)(x-3)

Apply the distributive property.

x⋅x2+x⋅x+x⋅1-1×2-1x-1⋅1(x+3)(x-3)

Reorder x and 1.

x⋅x2+x⋅x+1x-1×2-1x-1⋅1(x+3)(x-3)

Raise x to the power of 1.

x⋅x2+x⋅x+1x-1×2-1x-1⋅1(x+3)(x-3)

Use the power rule aman=am+n to combine exponents.

x1+2+x⋅x+1x-1×2-1x-1⋅1(x+3)(x-3)

Add 1 and 2.

x3+x⋅x+1x-1×2-1x-1⋅1(x+3)(x-3)

Raise x to the power of 1.

x3+x⋅x+1x-1×2-1x-1⋅1(x+3)(x-3)

Raise x to the power of 1.

x3+x⋅x+1x-1×2-1x-1⋅1(x+3)(x-3)

Use the power rule aman=am+n to combine exponents.

x3+x1+1+1x-1×2-1x-1⋅1(x+3)(x-3)

Add 1 and 1.

x3+x2+1x-1×2-1x-1⋅1(x+3)(x-3)

Multiply x by 1.

x3+x2+x-1×2-1x-1⋅1(x+3)(x-3)

Multiply -1 by 1.

x3+x2+x-1×2-1x-1(x+3)(x-3)

Move x.

x3+x2-1×2+x-1x-1(x+3)(x-3)

Subtract 1×2 from x2.

x3+0+x-1x-1(x+3)(x-3)

Add x3 and 0.

x3+x-1x-1(x+3)(x-3)

Subtract 1x from x.

x3+0-1(x+3)(x-3)

Add x3 and 0.

x3-1(x+3)(x-3)

x3-1(x+3)(x-3)

Expand (x+3)(x-3).

Apply the distributive property.

x3-1x(x-3)+3(x-3)

Apply the distributive property.

x3-1x⋅x+x⋅-3+3(x-3)

Apply the distributive property.

x3-1x⋅x+x⋅-3+3x+3⋅-3

Reorder x and -3.

x3-1x⋅x-3x+3x+3⋅-3

Raise x to the power of 1.

x3-1x⋅x-3x+3x+3⋅-3

Raise x to the power of 1.

x3-1x⋅x-3x+3x+3⋅-3

Use the power rule aman=am+n to combine exponents.

x3-1×1+1-3x+3x+3⋅-3

Add 1 and 1.

x3-1×2-3x+3x+3⋅-3

Multiply 3 by -3.

x3-1×2-3x+3x-9

Add -3x and 3x.

x3-1×2+0-9

Subtract 9 from 0.

x3-1×2-9

x3-1×2-9

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of 0.

x2 | + | 0x | – | 9 | x3 | + | 0x2 | + | 0x | – | 1 |

Divide the highest order term in the dividend x3 by the highest order term in divisor x2.

x | |||||||||||||

x2 | + | 0x | – | 9 | x3 | + | 0x2 | + | 0x | – | 1 |

Multiply the new quotient term by the divisor.

x | |||||||||||||

x2 | + | 0x | – | 9 | x3 | + | 0x2 | + | 0x | – | 1 | ||

+ | x3 | + | 0 | – | 9x |

The expression needs to be subtracted from the dividend, so change all the signs in x3+0-9x

x | |||||||||||||

x2 | + | 0x | – | 9 | x3 | + | 0x2 | + | 0x | – | 1 | ||

– | x3 | – | 0 | + | 9x |

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

x | |||||||||||||

x2 | + | 0x | – | 9 | x3 | + | 0x2 | + | 0x | – | 1 | ||

– | x3 | – | 0 | + | 9x | ||||||||

+ | 9x |

Pull the next term from the original dividend down into the current dividend.

x | |||||||||||||

x2 | + | 0x | – | 9 | x3 | + | 0x2 | + | 0x | – | 1 | ||

– | x3 | – | 0 | + | 9x | ||||||||

+ | 9x | – | 1 |

The final answer is the quotient plus the remainder over the divisor.

x+9x-1×2-9

The oblique asymptote is the polynomial portion of the long division result.

y=x

y=x

This is the set of all asymptotes.

Vertical Asymptotes: x=-3,3

No Horizontal Asymptotes

Oblique Asymptotes: y=x

Graph (x^3-1)/(x^2-9)