The rational function is .
The domain of a rational function is the set of all real numbers except those
\for which the denominator is .
Find which number make the fraction undefined create an equation where the
\denominator is not equal to .
.
The domain of the is the set of all real numbers
-except
.
The domain of function is
.
\
.
Rewrite the expression.
\.
is in lowest terms.
\
The rational function is .
Change to
.
.
Find the intercepts.
\To find -intercept equate numerator
.
.
Therefore there is no -intercept in given function
.
Find the -intercept by substituting
in the rational function.
- intercept is
.
\
Find the vertical asymptote by equating 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 zero and degree of the denominator is .
Since the degree of numerator is less than degree of denominator,
\so is a proper rational function, and the graph of
will have the
horizontal asymptote .
The function has horizontal asymptote at .
There are no zeros of numerator; the zeros of denominator are ,
use these values to divide the -axis into three intervals.
.
Interval | \ \
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Number chosen \ | \
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Value of | \
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Location of graph \ | \
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Above | \
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Below | \
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Above | \
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Point of graph \ | \
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End behavior of the graph:
\As and
, hence the graph of
approaches
to a vertical asymptote at .
As and
, hence the graph of
approaches
to a vertical asymptote at .
As and
, hence the graph of
approaches
to a horizontal asymptote at .
Graph :
\The graph of :
Step 1: ; Domain of function
is
.
Step 2: The function in lowest terms .
Step 3: There is no -intercepts and
-intercept is
.
Step 4: is in lowest terms.
Vertical asymptote at and
.
Step 5: Horizontal asymptote is ; not intersected.
\
Step 6:
\Interval | \ \
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Number chosen \ | \
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Value of | \
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Location of graph \ | \
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Above | \
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Below | \
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Above | \
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Point of graph \ | \
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Step 7 and step 8:
\Graph of :