y2-2y+12y-3

Find where the expression (y-1)22y-3 is undefined.

y=32

x=(y-1)22y-3 is an equation of a line, which means there are no horizontal asymptotes.

No Horizontal Asymptotes

Factor using the perfect square rule.

Rewrite 1 as 12.

y2-2y+122y-3

Check the middle term by multiplying 2ab and compare this result with the middle term in the original expression.

2ab=2⋅y⋅-1

Simplify.

2ab=-2y

Factor using the perfect square trinomial rule a2-2ab+b2=(a-b)2, where a=y and b=-1.

(y-1)22y-3

(y-1)22y-3

Expand (y-1)2.

Rewrite (y-1)2 as (y-1)(y-1).

(y-1)(y-1)2y-3

Apply the distributive property.

y(y-1)-1(y-1)2y-3

Apply the distributive property.

y⋅y+y⋅-1-1(y-1)2y-3

Apply the distributive property.

y⋅y+y⋅-1-1y-1⋅-12y-3

Reorder y and -1.

y⋅y-1y-1y-1⋅-12y-3

Raise y to the power of 1.

y⋅y-1y-1y-1⋅-12y-3

Raise y to the power of 1.

y⋅y-1y-1y-1⋅-12y-3

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

y1+1-1y-1y-1⋅-12y-3

Add 1 and 1.

y2-1y-1y-1⋅-12y-3

Multiply -1 by -1.

y2-1y-1y+12y-3

Subtract 1y from -1y.

y2-2y+12y-3

y2-2y+12y-3

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

2y | – | 3 | y2 | – | 2y | + | 1 |

Divide the highest order term in the dividend y2 by the highest order term in divisor 2y.

y2 | |||||||||

2y | – | 3 | y2 | – | 2y | + | 1 |

Multiply the new quotient term by the divisor.

y2 | |||||||||

2y | – | 3 | y2 | – | 2y | + | 1 | ||

+ | y2 | – | 3y2 |

The expression needs to be subtracted from the dividend, so change all the signs in y2-3y2

y2 | |||||||||

2y | – | 3 | y2 | – | 2y | + | 1 | ||

– | y2 | + | 3y2 |

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

y2 | |||||||||

2y | – | 3 | y2 | – | 2y | + | 1 | ||

– | y2 | + | 3y2 | ||||||

– | y2 |

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

y2 | |||||||||

2y | – | 3 | y2 | – | 2y | + | 1 | ||

– | y2 | + | 3y2 | ||||||

– | y2 | + | 1 |

Divide the highest order term in the dividend -y2 by the highest order term in divisor 2y.

y2 | – | 14 | |||||||

2y | – | 3 | y2 | – | 2y | + | 1 | ||

– | y2 | + | 3y2 | ||||||

– | y2 | + | 1 |

Multiply the new quotient term by the divisor.

y2 | – | 14 | |||||||

2y | – | 3 | y2 | – | 2y | + | 1 | ||

– | y2 | + | 3y2 | ||||||

– | y2 | + | 1 | ||||||

– | y2 | + | 34 |

The expression needs to be subtracted from the dividend, so change all the signs in -y2+34

y2 | – | 14 | |||||||

2y | – | 3 | y2 | – | 2y | + | 1 | ||

– | y2 | + | 3y2 | ||||||

– | y2 | + | 1 | ||||||

+ | y2 | – | 34 |

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

y2 | – | 14 | |||||||

2y | – | 3 | y2 | – | 2y | + | 1 | ||

– | y2 | + | 3y2 | ||||||

– | y2 | + | 1 | ||||||

+ | y2 | – | 34 | ||||||

+ | 14 |

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

y2-14+14(2y-3)

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

y=y2-14

y=y2-14

This is the set of all asymptotes.

Vertical Asymptotes: y=32

No Horizontal Asymptotes

Oblique Asymptotes: y=y2-14

Graph (y^2-2y+1)/(2y-3)