Example: y
= x + 1
x2 - 4
A. Find asymptotes
1.
Vertical: ( set denominator = 0 and solve for x
)
x = 2 and x = -2
2.
Horizontal: (as x gets larger, y
approaches
0)
y = 0
B. Does the function intersect its horizontal asymptote? (Set the function = to its horizontal asymptote value)
x + 1 = 0
x2 - 4
x + 1 = 0
x = -1 Yes, it does hit its asymptote, at (-1,0).
C. Plot points in each area separated by critical points (where y = 0 and where y is undefined)
x | y
3 | 4/5
0 |-1/4
-3/2
|
2/7
-3 |-2/5
D. Plot x- and y-intercepts: (0,-1/4) and (-1,0)
II. The highest power in the numerator = highest power in the denominator
Example: y
= x2 + 1
x2 - 9
A. Find asymptotes
1.
Vertical: ( set denominator = 0 and solve for x
)
x = 3 and x = -3
2.
Horizontal: (ratio of the leading coefficients)
y = 1
B. Does the function intersect its horizontal asymptote? (Set the function = to its horizontal asymptote value)
x2 + 1 =1
x2 - 9
x2 + 1 = x2 - 9
1 = -9 No, it does not hit its asymptote
OR ( divide and look at the remainder)
1 + 10/(x2 - 9)
x2 - 9 | x2 + 1
x2 - 9
10 Will 10 ever be 0?
No, so it does not hit its asymptote.
C. Plot points in each area separated by critical points (where y = 0 and where y is undefined)
x | y
4 | 17/7
0 |-1/9
-4 | 17/7 (also may use symmetry
around
the y-axis on this one.)
D. Plot x- and y-intercepts: (0, -1/9) There are no x-intercepts.
III. The highest power in the numerator > the highest power in the denominator.
Example:
y
= x2 - 1
x + 2
A. Find asymptotes
1.
Vertical: ( set denominator = 0 and solve for x
)
x = -2
2. Slant or Oblique: (divide numerator by denominator; quotient is the asymptote)
x - 2 + 3/(x + 2) As x gets
large,
3/(x
+ 2) approaches 0.
x + 2 | x2 + 0x - 1
x2 + 2x
So the asymptote would be y = x - 2.
-2x - 1
-2x - 4
3
B. Does the function intersect its slant/oblique asymptote? (Set the function = to its asymptote equation)
x2 - 1 = x - 2
x + 2
x2 - 1 = x2 - 4
1 = - 4 No, it does not hit its asymptote.
C. Plot points in each area separated by critical points (where y = 0 and where y is undefined)
x | y
0 |-1/2
-3 |-8
D. Plot x- and y-intercepts: (0, -1/2), (1,0), and (-1,0)
IV. Some of the factors cancel out
Then a hole will occur where
the canceled factor = 0.
Example:
y
= x2 - 1
x + 1
= (x + 1)(x - 1)
x + 1
= x - 1
This rational
equation is the same as y = x - 1 everywhere except
where
x
= -1.
The rational
equation is undefined there, while the line has a value of -2.
Graph the
line
y
= x - 1, drawing an open circle at (-1,-2),
because the rational
function
is not defined there.
Now to find the equation of a function for which we know some information:
Find an equation for a rational function with vertical asymptote x
= -2,
a slant asymptote y = x + 3, and a zero at x =
-4.
The fact that the function has
a slant asymptote, vertical asymptote, and a zero means
the function looks like (x + 4)(x - a) .
x + 2
The fact that the function has
a slant asymptote and a vertical asymptote means
the function looks like (x + 3) +
R .
x + 2
The two expressions
represent
the same function, so
(x
+ 4)(x - a) = (x + 3) + R
x + 2
x + 2
Getting a
common
denominator on the right side,
(x + 4)(x - a) = (x + 3)(x + 2) +
R
x + 2
x +
2
x + 2
(x + 4)(x - a) = x2 + 5x +
6
+ R
x + 2
x + 2
Multiplying
both sides by x + 2,
(x + 4)(x - a) = x2 + 5x + 6 + R
x2 + 4x - ax - 4a = x2 + 5x + 6 + R
x2 + (4 - a)x - 4a = x2 + 5x + (6 + R)
If the
expressions
are equal, the coefficient of x2 are the
same,
the coefficients
of x
are the same, and the constant terms are the same. So,
4 - a = 5 and - 4a = 6 + R
Solve the first equation for a and substitute it into the second equation.
-a =
1
- 4(-1) = 6 + R
a =
-1
4 = 6 + R
-2 = R
So, (x
+ 4)(x + 1) = x + 3 -
2
x +
2
x + 2
and the rational
function
can be written
y = (x + 4)(x + 1)
OR
y = x + 3 - 2
x +
2
x + 2
Problems (there are no questions here yet)
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