# How do you find vertical, horizontal and oblique asymptotes for f(x) = (x^2-4x+1) /( 2x-3)?

May 28, 2017

The vertical asymptote is $x = \frac{3}{2}$
The oblique asymptote is $y = \frac{1}{2} x - \frac{5}{4}$
No horizontal asymptote

#### Explanation:

As we cannot divide by $0$, $x \ne \frac{3}{2}$

The vertical asymptote is $x = \frac{3}{2}$

We perform a long division

$\textcolor{w h i t e}{a a a a}$$2 x - 3$$\textcolor{w h i t e}{a a a a}$$|$${x}^{2} - 4 x + 1$$\textcolor{w h i t e}{a a a a}$$|$$\frac{1}{2} x - \frac{5}{4}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$${x}^{2} - \frac{3}{2} x$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$$0 - \frac{5}{2} x + 1$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a}$$- \frac{5}{2} x - \frac{15}{4}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$$- 0 - \frac{11}{4}$

Therefore,

$f \left(x\right) = \frac{{x}^{2} - 4 x + 1}{2 x - 3} = \left(\frac{1}{2} x - \frac{5}{4}\right) - \frac{\frac{11}{4}}{2 x - 3}$

${\lim}_{x \to - \infty} f \left(x\right) - \left(\frac{1}{2} x - \frac{5}{4}\right) = {\lim}_{x \to - \infty} - \frac{\frac{11}{4}}{2 x - 3} = {0}^{+}$

${\lim}_{x \to + \infty} f \left(x\right) - \left(\frac{1}{2} x - \frac{5}{4}\right) = {\lim}_{x \to + \infty} - \frac{\frac{11}{4}}{2 x - 3} = {0}^{-}$

The oblique asymptote is $y = \frac{1}{2} x - \frac{5}{4}$
graph{(y-(x^2-4x+1)/(2x-3))(y-1/2x+5/4)(y-1000(x-3/2))=0 [-14.24, 14.24, -7.12, 7.12]}