Q. 26 - 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.   

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

Step-by-Step Solution

Answer

The roots of $f\left(x\right)$ are $x=-1,2.$ $f\left(x\right)$ are discontinuous at $x=2,-2.$ There are no horizontal asymptotes of $f\left(x\right).$ The vertical asymptotes of $f\left(x\right)$ are at $x=-2.$

Step 1. Given information

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

Step 2. Solution

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

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

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

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

                      $x=-1,2.$

The roots of the function are $x=-1,2.$ The function $f\left(x\right)$ are discontinuous at $x=2,-2.$ There are no horizontal asymptotes of $f\left(x\right).$ The value of $x$ that cause the denominator asymptotes are $x=-2.$