Chapter 12.1, Problem 67E

To determine

To calculate: The slant asymptote of the function $f\left(x\right)=\frac{x^2-4x+2}{x-3}$ .

Answer to Problem 67E

The slant asymptote is $y=x-1$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=\frac{x^2-4x+2}{x-3}$ .

Calculation:

Consider the function $f\left(x\right)=\frac{x^2-4x+2}{x-3}$ .

Apply the method of long division to divide the polynomial $x^2-4x+2$ by $x-3$ .

  $x-3\begin{array}{c}\hfill x-1\\ \hfill \overline{)\begin{array}{l}\begin{array}{ccccc}{x}^2& -& 4x& +& 2\end{array}\\ \begin{array}{ccccc}{x}^2& -& 3x& & \end{array}\\ \begin{array}{ccccc}& & -x& +& 2\end{array}\\ \begin{array}{ccccc}& & -x& +& 3\end{array}\\ \begin{array}{ccccc}& & -1& & \end{array}\end{array}}\end{array}$

Therefore, $\frac{x^2-4x+2}{x-3}=x-1+\frac{\left(-1\right)}{x-3}$ .

Observe that when $x\to \infty$ the value $\frac{\left(-1\right)}{x-3}\to 0$ . It is concluded that the graph of $f\left(x\right)=\frac{x^2-4x+2}{x-3}$ approach to the function $y=x-1$ .

Therefore, the function $y=x-1$ is the slant asymptote of the function $f\left(x\right)=\frac{x^2-4x+2}{x-3}$ .

Sketch the graph of the function $f\left(x\right)=\frac{x^2-4x+2}{x-3}$ and its slant asymptote to verify the result.

  

The red curve represent the function and blue line represent the slant asymptote.

Thus, the slant asymptote is $y=x-1$ .