Chapter 4 - il02222025.schoolwires.netβ¬Β¦Β Β· Chapter 4 Polynomial and ... Section 2 Properties of...
Transcript of Chapter 4 - il02222025.schoolwires.netβ¬Β¦Β Β· Chapter 4 Polynomial and ... Section 2 Properties of...
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Chapter 4Polynomial and Rational
Functions
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Section 2Properties of Rational Functions
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DAY 1
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A rational function has the form
R x =π(π₯)
π(π)
Where p(x) and q(x) are polynomials and q(x) β 0.
Domain: all real numbers except any x-values that would make q(x) = 0
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Example:Find the domain of each rational function
(set denominator = 0, solve for x domain excludes the solution)
R(x) = 4π₯
π₯β3
R(x) = π₯2βπ₯β6
4(π₯2β9)
R(x) = 3π₯2+π₯
π₯2+4
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Graphing Using Transformations of 1) 1
π₯2and 2)
1
π₯
What x-values would make each equation undefined?
What number could 1) and 2) not equal?b/c numerator is 1, the fraction willnever reduce to 0
(y-value)
Before we would use -1, 0, 1 to find basic points. Now we can only use -1 and 1 because x β 0
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Basics:*Vertical Asymptotes = xβs that make the function undefined*Horizontal Asymptotes = what graph approaches for large xβs
1
π₯21
π₯
(-1, 1) (1, 1) (-1, -1) (1, 1)V.A. x = 0; H.A. y = 0 V.A. x = 0; H.A. y = 0
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Example: Graphing Using Transformations
R(x) = 1
(π₯β2)2+ 1
1/x^2 (-1, 1) (1, 1); VA x = 0, HA y = 0
-2 right 2 (1, 1) (3, 1); VA x = 2, HA y = 0
1 (numerator) NO STRETCH/COMPRESS/REFLECT
+1 up 1 (1, 2) (3, 2); VA x = 2, HA y = 1
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Example: Graphing Using Transformations
R(x) = β2
(π₯+1)β 3
1/x (-1, -1) (1, 1); VA x = 0, HA y = 0
+1 left 1 (-2, -1) (0, 1); VA x = -1, HA y = 0
-2 S/R (-2, 2) (0, -2); VA x = -1, HA y = 0
-3 down 3 (-2, -1) (0, -5); VA x = -1, HA y = -3
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DAY 2
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Finding Vertical Asymptotes:
Vertical asymptotes represent x-values that would make the equation in lowest terms undefined
simplify = cancel like terms from top to bottom then set denominator = 0 and solve for x
**any terms that cancel out creates a βholeβ
The graph can never cross or touch a vertical asymptote
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Example:
R(x) = π₯
π₯2β4R(x) =
π₯2
π₯2+9R(x) =
π₯2β9
π₯2+4π₯β21
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Horizontal or oblique (a line on a diagonal y = mx + b)
Lines that represent where the graph is approaching for large values of x
It is possible to cross these asymptotes
Canβt have a horizontal and oblique at the same time
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To find them you have to compare the degree of the top polynomial to the degree of the bottom.
For each part, assume R(x) = π(π₯)
π(π₯)
1) If degree of p is less than degree of q H.A. y = 0
2) If degree of p is equal to degree of q H.A. y = πππππππ πππππ.ππ π‘ππ
πππππππ πππππ.ππ πππ‘π‘ππ
3) If degree on top is exactly one bigger than degree on bottom O.A. y = quotient
4) If none of the first 3 are true, there are no oblique or horizontal
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Examples:
1) degree of p < degree of q
R(x) = π₯
π₯3+1
2) degree of p = degree of q
R(x) = 6π₯2+π₯+2
4π₯2+1
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Examples:3) Degree of p > degree of q (by exactly one)
R(x) = 3π₯4βπ₯2
π₯3βπ₯2+1NEED TO DO LONG DIVISON!
CAN STOP WHEN DEGREE OF DIVIDEND IS SMALLER THAN DIVISOR
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Example:
4) None of the first 3 types are true
R(x) = π₯5+3π₯
π₯3+2π₯+1
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EXIT SLIP