$\"\"$

\

The rational function is $\"\"$ .

\

Find the intercepts:

\

The function is $\"\"$ .

\

Change $\"\"$ to $\"\"$ .

\

$\"\"$ .

\

Find $\"\"$ -intercept by equating the numerator to zero.

\

$\"\"$

\

$\"\"$ and $\"\"$

\

$\"\"$ and $\"\"$ .

\

The $\"\"$ -intercept is $\"\"$ and $\"\"$ .

\

Find $\"\"$ -intercept by substituting $\"\"$ in $\"\"$ .

\

$\"\"$

\

The $\"\"$ -intercept is $\"\"$ .

\

\

Find the vertical asymptotes :

\

Find the vertical asymptote by equating the denominator to zero.

\

$\"\"$

\

Thus, the function has vertical asymptote at $\"\"$ .

\

Find the horizontal asymptote :

\

To find horizontal asymptote, first find the degree of the numerator and degree of the denominator.

\

Degree of the numerator is $\"\"$ and degree of the denominator is $\"\"$ .

\

Since the degree of numerator is equal to the degree of the denominator, horizontal asymptote is ratio of leading coefficients.

\

The function has horizontal asymptote, at $\"\"$ .

\

\

Graph the function $\"\"$ .

\

Draw a coordinate plane.

\

Plot the intercepts and asymptotes.

\

Draw the curve.

\

Graph:

\

Graph of the function $\"\"$ .

\

$\"\"$

\

Find the domain:

\

Observe the graph of the function : The function is undefined at $\"\"$ and $\"\"$ .

\

Thus, the function is continuous for all real numbers except $\"\"$ and $\"\"$ .

\

Therefore, domain $\"\"$ .

\

$\"\"$

\

Horizontal asymptote at $\"\"$ .

\

Vertical asymptotes at $\"\"$ and $\"\"$ .

\

The $\"\"$ -intercepts are $\"\"$ and $\"\"$ .

\

The $\"\"$ -intercept is $\"\"$ . \ \ Graph of the function $\"\"$ :

\

$\"\"$

\

Domain : $\"\"$ .