Q. 27 - Calculus

Question

Find the roots, discontinuous, and horizontal and vertical asymptotes of the function in Exercises 23-24. Support your answers by explicitly computing any relevant limits.    

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

Step-by-Step Solution

Answer

The roots of $f\left(x\right)$ are $x=-1,2.$ The function is discontinuous at $x=2,-2.$ The horizontal asymptotes are at $y=0.$ The vertical asymptotes are at $x=-2,2.$

Step 1. Given information

The given function is $f\left(x\right)=\frac{\left(x+1\right)\left(x-2\right)}{{\left(x-2\right)}^2\left(x+2\right)}.$ We need to determine the roots, discontinuous, horizontal, and vertical asymptotes of $f\left(x\right).$

Step 2. Solution

$f\left(x\right)=\frac{\left(x+1\right)\left(x-2\right)}{{\left(x-2\right)}^2\left(x+2\right)}.$

Set $f\left(x\right)=0.$

$\frac{\left(x+1\right)\left(x-2\right)}{{\left(x-2\right)}^2\left(x+2\right)}=0$

     $\left(x+1\right)\left(x-2\right)=0.$

$x=-1,2.$

The roots of $f\left(x\right)$ are $x=-1,2.$ The function is discontinuous at x=2,-2.

By comparing the degree of the numerator and denominator the horizontal asymptotes are at $y=0.$ 

The value of $x$ that cause the denominator of $f\left(x\right)$ to zero are $-2,2.$ So, the vertical asymptotes are at $x=-2,2.$