find the horizontal and vertical asymptotes of the curve.\ny = \\frac{x^{2}-x}{x^{2}-5x + 4}\nx=\ny=

Explanation:

Step1: Find vertical asymptotes

Set the denominator equal to 0: $x^{2}-5x + 4=0$. Factor it as $(x - 1)(x - 4)=0$. Solving gives $x=1$ and $x = 4$.

Step2: Find horizontal asymptotes

Since the degrees of the numerator and denominator are the same (both degree 2), divide the leading - coefficients. The leading coefficient of the numerator is 1 and of the denominator is 1. So, $\lim_{x\rightarrow\pm\infty}\frac{x^{2}-x}{x^{2}-5x + 4}=\lim_{x\rightarrow\pm\infty}\frac{1-\frac{1}{x}}{1-\frac{5}{x}+\frac{4}{x^{2}}}=1$.

Answer:

$x = 1,x = 4$ $y = 1$