$\"\"$

The inequality is $\"\"$

The denominator of the function shoud not be zero.

$\"\"$

Therefore the domain of the inequality is $\"\"$ .

$\"\"$

Consider the function is $\"\"$ .

Find the $\"\"$ -intercept by susbtituting $\"\"$ .

$\"\"$

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

$\"\"$

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

$\"\"$

Vertical asymptote can be found by making denominator to zero.

$\"\"$

$\"\"$ .

The function has vertical asymptote at $\"\"$ .

$\"\"$

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 the numerator is equal to the degree of the denominator, horizontal asymptote is the ratio of leading coefficient of numerator and denominator.

Leading coefficient of numerator is $\"\"$ , leading coefficient of denominator is $\"\"$ .

$\"\"$ is horizontal asymptote.

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

$\"\"$

The zero of the numerator is $\"\"$ ;the zero of denominator is $\"\"$ ,use these values to divide the

$\"\"$ -axis into three intervals.

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

$\"\"$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

Interval	\ $\"\"$ \	\ $\"\"$ \	\ $\"\"$ \
\ Number chosen \	\ $\"\"$ \	\ $\"\"$ \	\ $\"\"$ \
\ Value of $\"\"$ \	\ $\"\"$ \	\ \ $\"\"$ \	\ $\"\"$ \
\ Location of graph \	\ above $\"\"$ -axis \	\ Below $\"\"$ -axis \	\ above $\"\"$ -axis \
\ Point of graph \	\ $\"\"$ \	\ $\"\"$ \	\ $\"\"$ \

$\"\"$

The function $\"\"$ is above the $\"\"$ - axis on the intervals $\"\"$ and $\"\"$ .

Therefore ,the solution set is $\"\"$ .

$\"\"$

The solution set is $\"\"$ .