holt mcdougal algebra 2 rational functions holt algebra 2 holes & slant asymptotes holes &...
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Holt McDougal Algebra 2
Rational FunctionsRational Functions
Holt Algebra 2
Holes & Slant AsymptotesHoles & Slant Asymptotes
Holt McDougal Algebra 2
Holt McDougal Algebra 2
Rational Functions
Now we find out why we have to start out factoring….to find holes in the graph.
Holt McDougal Algebra 2
Rational Functions
In some cases, both the numerator and the denominator of a rational function will equal 0 for a particular value of x. As a result, the function will be undefined at this x-value. If this is the case, the graph of the function may have a hole. A hole is an omitted point in a graph.
Holt McDougal Algebra 2
Rational Functions
Example : Graphing Rational Functions with Holes
(x – 3)(x + 3)x – 3
f(x) =
Identify holes in the graph of f(x) = . Then graph.
x2 – 9 x – 3
Factor the numerator.
The expression x – 3 is a factor of both the numerator and the denominator. Set it = to 0 to find the x- coordinate of the hole.
There is a hole in the graph at x = 3.
Divide out common factors.
(x – 3)(x + 3)(x – 3)
For x ≠ 3,
f(x) = = x + 3
Holt McDougal Algebra 2
Rational Functions
Example Continued
The graph of f is the same as the graph of y = x + 3, except for the hole at x = 3. On the graph, indicate the hole with an open circle. The domain of f is all real #’s except 3
Hole at x = (3, 6)
To find the y-coordinate of the hole plug 3 into the reduced equation.
Holt McDougal Algebra 2
Rational Functions
Check It Out! Example 5
(x – 2)(x + 3)x – 2
f(x) =
Identify holes in the graph of f(x) = . Then graph.
x2 + x – 6 x – 2
Factor the numerator.
The expression x – 2 is a factor of both the numerator and the denominator. Set it = to 0 to find the x- coordinate of the hole.
There is a hole in the graph at x = 2.
Divide out common factors.
For x ≠ 2,
f(x) = = x + 3(x – 2)(x + 3)(x – 2)
Holt McDougal Algebra 2
Rational Functions
Check It Out! Example 5 Continued
The graph of f is the same as the graph of y = x + 3, except for the hole at x = 2. On the graph, indicate the hole with an open circle. The domain of f is {x|x ≠ 2}.
Hole at x = 2
x 2
f xx 2 x 2
Domain: , 2 , 2, 2 , 2,
Vertical Asymptotes: x 2
Horizontal Asymptotes: y 0
Holes: 1
2,4
Intercepts:
1
x 2
10,
2
10,
2
2
2
x 5x 6f x
x 2x 3
Domain: , 1 , 1, 3 , 3,
Vertical Asymptotes: x 1
Horizontal Asymptotes: y 1
Holes: 1
3,4
Intercepts: 0, 2
2, 0
x 2 x 3
x 1 x 3
Domain: , 3 , 3, 3 , 3,
Vertical Asymptotes: x 3
Horizontal Asymptotes: y 0
Holes: 1
3,2
Intercepts: 0, 1
2
3x 9f x
x 9
3 x 3
x 3 x 3
3
x 3
Holt McDougal Algebra 2
Rational Functions
Let’s go back and look at the worksheet and find the problems with holes.
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 use long division – dividing the numerator by the denominator.
Holt McDougal Algebra 2
Rational Functions
When dividing to find slant asymptotes:
• Do synthetic division (if possible); if not, do long division!
• The resulting polynomial (ignoring the remainder) is the equation of the slant asymptote.
EXAMPLE: Finding the Slant Asymptoteof a Rational Function
Find the slant asymptotes of f (x) 2 4 5 .
3x x
x
Solution Because the degree of the numerator, 2, is exactly one more than the degree of the denominator, 1, the graph of f has a slant asymptote. To find the equation of the slant asymptote, divide x 3 into x2 4x 5:
2 1 4 51 3 3 1 1 8
3
2
81 13
3 4 5
xx
x x x
Remainder
Rational Functions and Their Graphs
moremore
EXAMPLE: Finding the Slant Asymptoteof a Rational Function
Find the slant asymptotes of f (x) 2 4 5 .
3x x
x
Solution The equation of the slant asymptote is y x 1. Using our
strategy for graphing rational functions, the graph of f (x) is
shown.
2 4 53
x xx
-2 -1 4 5 6 7 8321
7
6
5
4
3
1
2
-1
-3
-2
Vertical asymptote: x = 3
Vertical asymptote: x = 3
Slant asymptote: y = x - 1
Slant asymptote: y = x - 1
3.6: Rational Functions and Their Graphs
Graph: 22 3
( )1
x xf x
x
Notice that in this function, the degree of the numeratoris larger than the denominator. Thus n>m and there is nohorizontal asymptote. However, if n is one more than m,the rational function will have a slant asymptote.
To find the slant asymptote, divide the numerator by the denominator:
2
2
2 5
1 2 3
2 2
5
5 5
5
x
x x x
x x
x
x
The result is . We ignore the remainder and the line is a slant asymptote.
512 5 xx
2 5y x
Graph:
22 3
1
x x
x
1st, find the vertical asymptote.
2nd , find the x-intercepts: and
3rd , find the y-intercept:
4th , find the horizontal asymptote. none
0,0
0,0
3,0
2
1x
2 3
1
x x
x
5th , find the slant asymptote: 2 5y x
6th , sketch the graph.
A Rational Function with a Slant Asymptote
Graph the rational function
• Factoring:
2 4 5( )
3
x xr x
x
( 1)( 5)
( 3)
x xy
x
A Rational Function with a Slant Asymptote
Finding the x-intercepts:
• –1 and 5 (from x + 1 = 0 and x – 5 = 0)
Finding the y-intercepts:
• 5/3 (because )20 4 0 5 5
(0)0 3 3
r
A Rational Function with a Slant Asymptote
Finding the horizontal asymptote:• None (because degree of numerator is greater
than degree of denominator)
Finding the vertical asymptote:• x = 3 (from the zero of the denominator)
A Rational Function with a Slant Asymptote
Finding the slant asymptote:
• Since the degree of the numerator is one more than the degree of the denominator, the function has a slant asymptote.
• Dividing, we obtain:
• Thus, y = x – 1 is the slant asymptote.
8( ) 1
3r x x
x
A Rational Function with a Slant Asymptote
Here are additional values and
the graph.
So far, we have considered only horizontal
and slant asymptotes as end behaviors for
rational functions.
• In the next example, we graph a function whose end behavior is like that of a parabola.
Slant Asymptotes and End Behavior
Finding a Slant Asymptote
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 AsymptoteCon’t.
39521 232
xxxxxx
xxx 23
943 2 xx
333 2 xx
127 x
Finding a Slant AsymptoteCon’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
3x3xy
Graph of Example 7
The slanted line y = x + 3 is the slant asymptote
Graph the rational function which has the following characteristics
Vert Asymp at x = 1, x = -3
Horz Asymp at y = 1
Intercepts (-2, 0), (3, 0), (0, 2)
Passes through (-5, 2)
Graph the rational function which has the following characteristics
Vert Asymp at x = 1, x = -1
Horz Asymp at y = 0
Intercepts (0, 0)
Passes through (-0.7, 1), (0.7, -1), (-2, -0.5), (2, 0.5)
Holt McDougal Algebra 2
Rational Functions
zero: (2,0); asymptotes: x = 0, y = 1; hole at (1, -1)
Identify the zeros, asymptotes, and holes in the graph of . Then graph. x2 – 3x + 2
x2 – x f(x) =