91Ó°ÊÓ

Given to determine the roots, discontinuities, and horizontal and vertical asymptotes of the functions below.

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

The roots of a function are the values of x where the simplified function value is 0.

Simplifying the function and finding the root:

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

So the root of the function is -1.

A function has a discontinuity at a point $x=a$where $f\left(a\right)≠\underset{x→a}{\lim }f\left(x\right)$.

The function is undefined when the denominator is 0.

Here,

$$\begin{gathered} \left(x-2\right){\left(x+2\right)}^2=0 \\ x-2=0,x+2=0 \\ x=2,x=-2 \end{gathered}$$

So the function has discontinuity at$x=-2,2$

Here the degree of the denominator is greater than the degree of the numerator. Hence the horizontal asymptote is$y=0$.

The vertical asymptote of a rational function are the zeroes of the denominator of the simplified function.

Here the simplified function is:

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

When the denominator is 0:

$$\begin{gathered} {\left(x+2\right)}^2=0 \\ x+2=0 \\ x=-2 \end{gathered}$$

So the function has a vertical asymptote at$x=-2$

The root of the function is $x=-1$.

The function has discontinuities at $x=-2,2$.

The function has a horizontal asymptote at $y=0$.

The function has a vertical asymptote at$x=-2$.