Our restriction here is that the denominator of a
fraction can never
be equal to 0. So to find our domain, we want to set the
denominator
equal to 0 and restrict those values.

Our domain is all real
numbers except 2 and -5, because 2 and -5 both make the
denominator equal to 0, which would not give us a real number answer
for our function.

Reducing the rational
function to lowest terms we get:

Vertical Asymptote:

So now we want to find where the denominator is equal to 0:

So now we want to find where the denominator is equal to 0:

There is one** vertical asymptote: ***x* = -1/3.

Horizontal Asymptote:

So now we want to compare the degrees of the numerator and the denominator.

What is the degree of the numerator that is left? If you said 0, you are correct. The leading term is the constant 1 and its degree is 0.

What is the degree of the denominator that is left? If you said 1, you are correct. The leading term is x and its degree is 1.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is 1 is less than the degree of the denominator, then there is a horizontal asymptote at y = 0.

So now we want to compare the degrees of the numerator and the denominator.

What is the degree of the numerator that is left? If you said 0, you are correct. The leading term is the constant 1 and its degree is 0.

What is the degree of the denominator that is left? If you said 1, you are correct. The leading term is x and its degree is 1.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is 1 is less than the degree of the denominator, then there is a horizontal asymptote at y = 0.

This rational function
is already in lowest terms:

Vertical Asymptote:

So now we want to find where the denominator is equal to 0:

So now we want to find where the denominator is equal to 0:

There are two** vertical asymptotes: ***x* = -1/2 and 1/2.

Horizontal Asymptote:

So now we want to compare the degrees of the numerator and the denominator.

What is the degree of the numerator? If you said 2, you are correct. The leading term is the constant and its degree is 2.

What is the degree of the denominator? If you said 2, you are correct. The leading term is and its degree is 2.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at

So now we want to compare the degrees of the numerator and the denominator.

What is the degree of the numerator? If you said 2, you are correct. The leading term is the constant and its degree is 2.

What is the degree of the denominator? If you said 2, you are correct. The leading term is and its degree is 2.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at

**Note that this rational function is already reduced
down.
**

Applying long
division to this problem we get:

The answer to the long division would be .

The equation for the slant asymptote is the quotient part of the answer which would be .

The equation for the slant asymptote is the quotient part of the answer which would be .

This function cannot be reduced any further. This means that there will be no open holes on this graph.

Vertical Asymptote:

So now we want to find where the denominator is equal to 0:

So now we want to find where the denominator is equal to 0:

There are two** vertical asymptotes: ***x* = -3 and x = 3.

Horizontal Asymptote:

So now we want to compare the degrees of the numerator and the denominator.

What is the degree of the numerator? If you said 1, you are correct. The leading term is x and its degree is 1.

What is the degree of the denominator? If you said 2, you are correct. The leading term is and its degree is 2.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is 1 is less than the degree of the denominator, then there is a horizontal asymptote at y = 0.

So now we want to compare the degrees of the numerator and the denominator.

What is the degree of the numerator? If you said 1, you are correct. The leading term is x and its degree is 1.

What is the degree of the denominator? If you said 2, you are correct. The leading term is and its degree is 2.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is 1 is less than the degree of the denominator, then there is a horizontal asymptote at y = 0.

Slant Asymptote:

Since the degree of the numerator is NOT one degree higher than the degree of the denominator, there is not slant asymptote.

Since the degree of the numerator is NOT one degree higher than the degree of the denominator, there is not slant asymptote.

Since f(-x) = -f(x) , the function is odd.

This means the graph is symmetric about the origin.

If you need a review of even and odd functions, feel free to go to Tutorial 32: Graphs of Functions Part II.

x-intercept:

What value are we going to use for y? You are correct if you said 0.

What value are we going to use for y? You are correct if you said 0.

There is an x-intercept at (0, 0).

y-intercept:

What value are we going to use for x? You are correct if you said 0.

What value are we going to use for x? You are correct if you said 0.

There is an y-intercept at (0, 0).

Note how the vertical asymptotes sections the graph
into
three parts. I’m going to plug in two x values that are to the left
of x = -3, two in
between x = -3 and x = 3,
and two
that are to the right of x = 3.

Plugging in -5 for x we get:

(-5, 5/16)

(-5, 5/16)

Plugging in -4 for x we get:

(-4, 4/7)

(-4, 4/7)

Plugging in -1 for x we get:

(-1, -1/8)

(-1, -1/8)

Plugging in 1 for x we get:

(1, 1/8)

(1, 1/8)

Plugging in 4 for x we get:

(4, -4/7)

(4, -4/7)

Plugging in 5 for x we get:

(5, -5/16)

(5, -5/16)

The factor of x - 3 canceled out and there were no other factors of *x* - 3 left in the denominator. This means there is an open hole on
the graph at x = 3.

x = 3:

x = 3:

There is an open hole
at (3, 9/2).

Vertical Asymptote:

So now we want to find where the new denominator is equal to 0:

So now we want to find where the new denominator is equal to 0:

There is one** vertical asymptote: ***x* = -3.

Horizontal Asymptote:

Since the degree of the numerator is one degree higher than the degree of the denominator, there is a slant asymptote and no horizontal asymptote.

Since the degree of the numerator is one degree higher than the degree of the denominator, there is a slant asymptote and no horizontal asymptote.

Slant Asymptote:

**Applying long
division to this problem we get:**

The answer to the long division would be .

The equation for the slant asymptote is the quotient part of the answer which would be .

The equation for the slant asymptote is the quotient part of the answer which would be .

Since , the function is neither even nor odd.

If you need a review of even and odd functions, feel free to go to Tutorial 32: Graphs of Functions Part II.

x-intercept:

What value are we going to use for y? You are correct if you said 0.

What value are we going to use for y? You are correct if you said 0.

There are no real
number solutions to this equation, so there are no x-intercepts.

y-intercept:

What value are we going to use for x? You are correct if you said 0.

What value are we going to use for x? You are correct if you said 0.

The y-intercept is (0, 3).

Note how the
vertical asymptote sections the graph into two parts. I’m going
to
plug in two x values that are to the left of x = -3 and two x values that are to the right of x = -3 .

Plugging in -5 for x we get:

(-5, -19/2)

(-5, -19/2)

Plugging in -4 for x we get:

(-4, -13)

(-4, -13)

Plugging in -2 for x we get:

(-2, 7)

(-2, 7)

Plugging in -1 for x we get:

(-1, 7/2)

(-1, 7/2)

Last revised on March 17, 2011 by Kim Seward.

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