holt mcdougal algebra 2 rational functions holt algebra 2 holes & slant asymptotes holes &...
TRANSCRIPT
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Holt McDougal Algebra 2
Rational FunctionsRational Functions
Holt Algebra 2
Holes & Slant AsymptotesHoles & Slant Asymptotes
Holt McDougal Algebra 2
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Holt McDougal Algebra 2
Rational Functions
Now we find out why we have to start out factoring….to find holes in the graph.
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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.
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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
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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.
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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)
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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
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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
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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
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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
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Holt McDougal Algebra 2
Rational Functions
Let’s go back and look at the worksheet and find the problems with holes.
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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.
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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.
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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
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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
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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
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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.
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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
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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
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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)
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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
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A Rational Function with a Slant Asymptote
Here are additional values and
the graph.
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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
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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
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Finding a Slant AsymptoteCon’t.
39521 232
xxxxxx
xxx 23
943 2 xx
333 2 xx
127 x
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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
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Graph of Example 7
The slanted line y = x + 3 is the slant asymptote
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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)
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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)
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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) =