Question

Write a rational function $f$ that has vertical asymptote at $x=2$, a horizontal asymptote at $y=3$ and a zero at $x=-1$.

• A. $f\left(x\right)=\frac{3(x-5)}{(x+2)}$
• B. $f\left(x\right)=\frac{3(x+1)}{(x-2)}$
  
• C. $f\left(x\right)=\frac{3(x-5)}{(x-2)}$
  
• D. $f\left(x\right)=\frac{3(x+5)}{(x+2)}$
  

Answer 1

The correct option is B $f\left(x\right)=\frac{3(x+1)}{(x-2)}$

$y=\frac{f\left(x\right)}{g\left(x\right)}$

Vertical asymptode at $r=2$

$y⟶\infty ,x=2$

$y=\frac{f\left(x\right)}{x-2}$

Zero at $x=-1$

$y=\frac{a(x+1)}{x-2}$

$xy-2y=a+2y$

$x=\frac{a+2y}{y-a}$

$y\to 3,x\to \infty$

$a=3$

$y=\frac{3(x+1)}{x-2}$