# How do you find the vertical, horizontal or slant asymptotes for #f(x) = ( x^2-2x)/(x^2-5x+4)#?

##### 2 Answers

#### Answer:

vertical asymptotes at x = 1 , x = 4

horizontal asymptote at y = 1

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

solve

#x^2-5x+4 = 0 → (x-1)(x-4) = 0 → x= 1 , x=4# Horizontal asymptotes occur as

#lim_(x→±∞) f(x) → 0# If the degree of the numerator and denominator are equal , as in this case , both degree 2, then the equation can be found by taking the ratio of leading coefficients.

#rArr y = 1/1 = 1 rArr y = 1 " is the equation "# here is the graph of the function.

graph{(x^2-2x)/(x^2-5x+4) [-10, 10, -5, 5]}

#### Answer:

There are two vertical asymptotes :

Horizontal asymptote :

Slant asymptote : None

#### Explanation:

For the Vertical Asymptote:

the factors of the denominator are

For the Horizontal Asymptote:

take the limiting value of the function as x approaches infinity

therefore

See the graph of

graph{y=(x^2-2x)/(x^2-5x+4)[-20,20,-10,10]}

See the graph of

graph{(y-10000x+10000)(y-10000x+4*10000)(y-0*x-1)=0[-20,20,-10,10]}

God bless...I hope the explanation is useful