Q. 23 - 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.

$f\left(x\right)=\frac{x^2-2x-3}{x-3}$

Step-by-Step Solution

Answer

The root of $f\left(x\right)$ is $0.$ $f\left(x\right)$ discontinuous at $x=3.$ $f\left(x\right)$ does not have any horizontal asymptotes. $f\left(x\right)$ has vertical asymptotes at $x=3.$

Step 1. Given information

The given function is $f\left(x\right)=\frac{x^2-2x-3}{x-3}.$ We need to determine the roots, discontinuous, and horizontal, and vertical asymptotes of the function $f\left(x\right).$

Step 2. Solution

$f\left(x\right)=\frac{x^2-2x-3}{x-3}.$

       $=\frac{x^2-3x+x-3}{x-3}.$

       $=\frac{x\left(x-3\right)+1\left(x-3\right)}{\left(x-3\right)}.$

        $=\frac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)}.$

The root of the function is $x=-1$ and the $f\left(x\right)$ is discontinuous at $x=3.$ To determine the horizontal asymptotes we must examine limit as $x\to \pm \infty .$

$\underset{x\to \infty }{\lim }f\left(x\right)=\frac{\left(x-3\right)\left(x+1\right)}{\left(x-3\right)}.$

The function $f\left(x\right)$ does not have any horizontal asymptote.

The value of $x$ that cause the denominator of $f\left(x\right)$ to zero is $3.$

So, $\underset{x\to \infty }{\lim }f\left(x\right)=\frac{\left(x-3\right)\left(x+1\right)}{\left(x-3\right)}.$

          $\underset{x\to 3}{\lim }f\left(x\right)=x+1.$

                        $=4.$

So, the vertical asymptotes at $x=3.$