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

Mar 2, 2016

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.

$\Rightarrow y = \frac{1}{1} = 1 \Rightarrow y = 1 \text{ is the equation }$

here is the graph of the function.
graph{(x^2-2x)/(x^2-5x+4) [-10, 10, -5, 5]}

There are two vertical asymptotes : $x = 1$ and $x = 4$
Horizontal asymptote : $y = 1$
Slant asymptote : None

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} - 2 x}{{x}^{2} - 5 x + 4}$

For the Vertical Asymptote:
the factors of the denominator are
$\left(x - 4\right) \left(x - 1\right)$ so that the vertical asymptotes are
$x = 1$ and $x = 4$

For the Horizontal Asymptote:

take the limiting value of the function as x approaches infinity

${\lim}_{x \rightarrow \infty} \frac{{x}^{2} - 2 x}{{x}^{2} - 5 x + 4} = {\lim}_{x \rightarrow \infty} \frac{1 - \frac{2}{x}}{1 - \frac{5}{x} + \frac{4}{x} ^ 2} = 1$

therefore $y = 1$ is a horizontal asymptote

See the graph of $f \left(x\right) = \frac{{x}^{2} - 2 x}{{x}^{2} - 5 x + 4}$

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

See the graph of $x = 1$ and $x = 4$ and $y = 1$

graph{(y-10000x+10000)(y-10000x+410000)(y-0x-1)=0[-20,20,-10,10]}

God bless...I hope the explanation is useful