ASYMPTOTES TUTORIAL Horizontal Vertical Slant and Holes Do Now: graph the following and name the...
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Transcript of ASYMPTOTES TUTORIAL Horizontal Vertical Slant and Holes Do Now: graph the following and name the...
ASYMPTOTES TUTORIAL
HorizontalVertical
Slantand Holes
Do Now: graph the following and name the asymptotes:
y =1
x−4
Definition of an asymptote
An asymptote is a straight line which acts as a boundary for the graph of a function.
When a function has an asymptote (and not all functions have them) the function gets closer and closer to the asymptote as the input value to the function approaches either a specific value a or positive or negative infinity.
The functions most likely to have asymptotes are rational functions
Vertical Asymptotes
Vertical asymptotes occur when thefollowing condition is met: The denominator of the simplified rational function is equal to 0. Remember, the simplified rational function has cancelled any factors common to both the numerator and denominator.
Finding Vertical AsymptotesExample 1
Given the function
The first step is to cancel any factors common to both numerator and denominator. In this case there are none.
The second step is to see where the denominator of the simplified function equals 0.
x
xxf
22
52
1
01
012
022
x
x
x
x
Finding Vertical Asymptotes Example 1 Con’t.
The vertical line x = -1 is the only vertical asymptote for the function. As the input value x to this function gets closer and closer to -1 the function itself looks and acts more and more like the vertical line
x = -1.
Graph of Example 1
The vertical dotted line at x = –1 is the vertical asymptote.
Rational Functions and Asymptotes
Arrow notation:
€
x → a+
x → a−
x →+∞
x → −∞
Meaning:
x approaches a from the right
x approaches a from the left
x is approaching infinity, increasing forever
x is approaching infinity, decreasing forever
Vertical Asymptotes
The line x = a is a vertical asymptote of a function if f(x) increases or decreases without bound as x approaches a.
If
Then:
€
x → a+ or x → a−
€
f x( ) →+∞ or f x( ) → −∞
Graph of Example 1
The vertical dotted line at x = –1 is the vertical asymptote.
limx→−1+
=∞
limx→−1−
=−∞
Finding Vertical AsymptotesExample 2
If
First simplify the
function. Factor
both numerator
and denominator
and cancel any
common factors.
9
121022
2
x
xxxf
2x2 +10x+12x2 −9
=x+ 3( ) 2x+ 4( )x+ 3( ) x−3( )
=2x+ 4x−3
*There is no asymptote at x = -3!
Finding Vertical Asymptotes Example 2 Con’t.
The asymptote(s) occur where the simplified denominator equals 0.
The vertical line x=3 is the only vertical asymptote for this function.
As the input value x to this function gets closer and closer to 3 the function itself looks more and more like the vertical line x=3.
3
03
x
x
Graph of Example 2
The vertical dotted line at x = 3 is the vertical asymptote
limx→ 3+
=∞
limx→ 3−
=−∞
Finding Vertical AsymptotesExample 3
If
Factor both the numerator and denominator and cancel any common factors.
6
52
xx
xxg
Finding Vertical AsymptotesExample 3
If
Factor both the numerator and denominator and cancel any common factors.In this case there are no common factors to cancel.
6
52
xx
xxg
32
5
6
52
xx
x
xx
x
Finding Vertical AsymptotesExample 3 Con’t.The denominator equals zero whenever
either
or
This function has two vertical asymptotes, one at x = -2 and the other at x = 3
2
02
x
x
3
03
x
x
Graph of Example 3
The two vertical dotted lines at x = -2 and x = 3 are the vertical asymptotes
Write 2 limit notations for each one.
Horizontal asymptotes occur when either one of the following conditions is met (you should notice that both conditions cannot be true for the same function).Horizontal asymptotes guide the end behavior of graph.
If the degree of the numerator < degree of the denominator then the horizontal asymptote will be y= 0.
If the degree of the numerator = degree of denominator then the horizontal asymptote will be
If the degree of the numerator > degree of the denominator then there is no horizontal asymptote.
Find the horizontal asymptote:
1. f x( ) =
2x−1x+ 2
x. f x
x
3
2
12
H.A. : y 2
H.A. : none
H.A. : y 0
Exponents are the same; divide the coefficients
Bigger on Top; None
Bigger on Bottom; y=0
Finding Horizontal AsymptotesExample 4
If
then there is a horizontal asymptote at the line y=0 because the degree of the numerator (2) is less than the degree of the denominator (3). This means that as x gets larger and larger in both the positive and negative directions (x → ∞ and x → -∞)
the function itself looks more and more like the horizontal line y = 0
27
533
2
x
xxxf
Graph of Example 4
The horizontal line y = 0 is the horizontal asymptote.
limx→∞
=0 and limx→−∞
=0
Finding Horizontal Asymptotes Example 5
If
then because the degree of the numerator (2) is equal to the degree of the denominator (2)
there is a horizontal asymptote at the line y=6/5.
Note, 6 is the leading coefficient of the numerator and 5 is the leading coefficient of the denominator. As x→∞ and as x→-∞ g(x) looks
more and more like the line y=6/5
975
5362
2
xx
xxxg
Graph of Example 5
The horizontal dotted line at y = 6/5 is the
horizontal asymptote.
limx→∞
=65and lim
x→−∞=65
Finding End behavior Asymptotes Do Now: graph:
If
There are no horizontal asymptotes because the degree of the numerator is greater than the degree of the denominator.
1
9522
3
x
xxxf
Graph of previous function
Slant Asymptotes
Slant asymptotes occur when the degree of the numerator is exactly one bigger than the degree of the denominator. In this case a slanted line (not horizontal and not vertical) is the function’s asymptote.
To find the equation of the asymptote we need to using synthetic division, (if possible) or long division – dividing the numerator by the denominator.
Finding slant asymptotes with synthetic division:
f (x) =2x2 −3x+5
x+2
This one has a linear denominator so we can use synthetic division
Note: slants and horizontals don’t happen together!
Finding slant asymptotes with synthetic division is only possible when the denominator is linear!
f (x) =2x2 −3x+5
x+2
y =2x−7(Remainder was 19)
End behavior Asymptotes
Graph the following function:
The end behavior asymptote will be a quadraticfunction because the numerator is 2 degreeslarger in the numerator
Use synthetic division to find it.
f (x) =x3 −2xx+1
End behavior Asymptotes
Graph the following function:
Note that the remainders don’t matter here!
f (x) =x3 −2xx+1
y =x2 −x−1
Using long division
Find the quotient and remainder:
1
9522
3
x
xxxf
The slant asymptote
We divide: note that here we will put the first answer over the 5x.
1
9522
3
x
xxxf
The slant asymptote
We divide and find it….
1
9522
3
x
xxxf
x2 +1 −2x3 +0x2 +5x−9
(−2x3
−2x
−2x )
(7x −9)
The slant asymptote
We divide and find it….Y = -2x is the slant asymptote!
1
9522
3
x
xxxf
x2 +1 −2x3 +0x2 +5x−9
(−2x3
−2x
−2x )
(7x −9)
7x-9 is just a remainder
Graph of Example 6
Y=-2x
Finding a Slant Asymptote Example 7
If
There will be a slant asymptote because the degree of the numerator (3) is one bigger than the degree of the denominator (2).
Using long division, divide the numerator by the denominator.
1
9522
23
xx
xxxxf
Finding a Slant AsymptoteExample 7 Con’t.
xxx 23
943 2 xx
333 2 xx
127 x
Finding a Slant AsymptoteExample 7 Con’t.
We can ignore the remainder
The answer we are looking for is the quotient
and the equation of the slant asymptote is:
127 x
3x
3xy
Graph of Example 7
The slanted line y = x + 3 is the slant asymptote
Finding slant asymptotes with synthetic division:
f (x) =2x2 −3x+5
x+2
divide using synthetic division
Finding slant asymptotes with synthetic division:
f (x) =2x2 −3x+5
x+2
divide using long division
Finding slant asymptotes with synthetic division is only possible when the denominator is linear!
f (x) =2x2 −3x+5
x+2
y =2x−7(Remainder was 19)
Finding Intercepts of Rational Functions
x-intercept: solve f (x) = 0
y-intercept: evaluate f (0)
Do now: find the x and y intercepts:y=2x+8
Intercepts of Rational Functions
Example
x-intercept: solve f (x) = 0
.352
1)(
2
xxx
xf
.1 isintercept - The 1
01
0352
12
xx
xxx
x
Setting the function = 0 is essentially setting the numerator= to 0!
Intercepts of Rational Functions
y-intercept: evaluate f (0)
31
isintercept - The 31
3)0(5)0(210
)0( 2
yf
Find all asymptotes and intercepts
Example Graph
Vertical Asymptote:
Horizontal Asymptote:
x-intercept:
y-intercept:
.312
)(
xx
xf
Find the asymptotes and intercepts
Example
Solution Vertical Asymptote:
Horizontal Asymptote:
x-intercept:
y-intercept:
.312
)(
xx
xf
303 xx
212 yy
21or 01203
12 xxx
x
31
301)0(2
)0( f
Graph all the asymptotes and interceptsx=3, y=2 (-.5,0) (0, -1/3)
.312
)(
xx
xf
Graph all the asymptotes and interceptsx=3, y=2 (-.5,0) (0, -1/3)
.312
)(
xx
xf
But find a point for themissing part-try f(6)
Graph all the asymptotes and interceptsx=3, y=2 (-.5,0) (0, -1/3)
.312
)(
xx
xf
But find a point for themissing part-try f(6)
f (6) =2(6)+16−3
=133=4
13
End behavior:
Use long division and divide to find the slant asymptote:
Do Now:
Use long division and divide:
x3 −x2 2 x4 0x3 −2x2 3x6−(x4 − x3 +2x)
x3 −2x2 + x+6
x +1
−(x3 − x2 2)
−x2 + x + 4 remainder
Slant asymptote:Y = x + 1
Quadratic and cubic end behavior asymptotes….Use long division to find the quotient:
f (x) x4 2x3 −2x1x2 −3x 4
Quadratic and cubic end behavior asymptotes….Use long division to find the quotient:
Quadratic end behavior asymptote:
(Note: the remainder was 11x-43)
f (x) x4 2x3 −2x1x2 −3x 4
y =x2 +5x+11
do now:find the asymptotes
Holes occur in the graph of a rational function whenever the numerator and denominator have common factors. The holes occur at the x value(s) that make the common factors equal to 0.
The hole is known as a removable discontinuity.When you graph the function on your calculator you
won’t be able to see the hole but the function is still discontinuous.
f x( ) =2x2 +10x+12
x2 −9
Finding a Hole
We were able to cancel the (x + 3) in the numerator and denominator before finding the vertical asymptote.
Because (x + 3) is a common factor there will be a hole at the point where
Finding the coordinates of the hole: substitute -3 into the remaining factors:
f x( ) =2x2 +10x+12
x2 −9=
x+ 3( ) 2x+ 4( )x+ 3( ) x−3( )
3
03
x
x
2x + 4x−3
=2(−3)+ 4−3−3
=13
Finding a Hole
There is a hole at
Where are the intercepts and asymptotes?
f x( ) =2x2 +10x+12
x2 −9=
x+ 3( ) 2x+ 4( )x+ 3( ) x−3( )
(−3,13)
Finding a Hole
is a hole
(-2,0),and are intercepts
x=3, y=2 are the asymptotes
Sketch the function
f x( ) =2x2 +10x+12
x2 −9=
x+ 3( ) 2x+ 4( )x+ 3( ) x−3( )
(0,−43)
(−3,13)
Graph of Example
Notice there is a hole in the graph at the point where x = -3. You would not be able to see this hole if you graphed the curve on your calculator (but it’s there just the same.)
limx→ 3+
=∞
limx→ 3_
=−∞
Finding a Hole
Use the rational root theorem to factor the numerator
Factor both numerator and denominator to see if there are any common factors.
Find all the intercepts, holes and asymptotes toSketch the function
4
82
3
x
xxf
Finding a Hole
Factor both numerator and denominator to see if there are any common factors.
there will be a hole at (2,3) asymptotes are x=-2And a slant at y = x (long division) and an
intercept at (0,2). No x intercepts.
22
422
4
8 2
2
3
xx
xxx
x
xxf
4
82
3
x
xxf
Graph of last example
There is a hole in the curve at the point where x = 2. This curve also has a vertical asymptote at x = -2 and a slant asymptote y = x.
ProblemsFind the vertical asymptotes, horizontal asymptotes, slant asymptotes and holes and intercepts for each of the following functions. Not all will be found for each. Graph g(x)
2
2
2 15
7 10
x xf x
x x
Vertical: Horizontal : Slant: Hole:
22 5 7
3
x xg x
x
Vertical: Horizontal : Slant: Hole:
Problems
2
2
2 15
7 10
x xf x
x x
Vertical: x = -2Horizontal : y = 1Slant: noneHole: at x = - 5x int -3y int -.5
22 5 7
3
x xg x
x
Vertical: x = 3Horizontal : noneSlant: y = 2x +11Hole: nonex int: -3.5,1y int: 7/3
Graph of g(x) 22 5 7
3
x xg x
x
Vertical: x = 3Horizontal : noneSlant: y = 2x +11Hole: none int:
-3.5,1y int: 7/3
Graphing a rational function.
1. Find the vertical asymptotes if they exist.2. Find the horizontal asymptote if it exists.3. Find the x intercept(s) by finding f(x) = 0.4. Find the y intercept by finding f(0).5. Find any crossing points on the horizontal
asymptote by setting f(x) = the horizontal asymptote.
6. Graph the asymptotes, intercepts, and crossing points.
7. Find additional points in each section of the graph as needed.
8. Make a smooth curve in each section through known points and following asymptotes and end behavior.