the slant asymptote of $f(x)=\frac{x^{2}-4x + 1}{2x - 3}$ is $y=\frac{1}{2}x-\frac{5}{4}$. please select the…

Explanation:

Step1: Perform polynomial long - division

Divide $x^{2}-4x + 1$ by $2x-3$. We know that when we divide a polynomial $f(x)=a_nx^n+\cdots+a_0$ by $d(x)=b_mx^m+\cdots + b_0$ ($n\geq m$), for a slant asymptote we consider the quotient of the long - division. Using polynomial long - division: [ \begin{align*} \frac{x^{2}-4x + 1}{2x-3}&=\frac{\frac{1}{2}x(2x - 3)-\frac{3}{2}x-4x + 1}{2x-3}\ &=\frac{\frac{1}{2}x(2x - 3)-\frac{11}{2}x + 1}{2x-3}\ &=\frac{1}{2}x+\frac{-\frac{11}{2}x + 1}{2x-3}\ &=\frac{1}{2}x-\frac{5}{4}+\frac{-\frac{7}{4}}{2x - 3} \end{align*} ] As $x\to\pm\infty$, the term $\frac{-\frac{7}{4}}{2x - 3}\to0$. The quotient of the long - division is $\frac{1}{2}x-\frac{5}{4}$, which is the equation of the slant asymptote.

Answer:

A. T