Post on 12-Aug-2020
Chapter 4Polynomial and Rational
Functions
Section 2Properties of Rational Functions
DAY 1
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
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
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
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
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
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
DAY 2
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
Example:
R(x) = π₯
π₯2β4R(x) =
π₯2
π₯2+9R(x) =
π₯2β9
π₯2+4π₯β21
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
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
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
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
Example:
4) None of the first 3 types are true
R(x) = π₯5+3π₯
π₯3+2π₯+1
EXIT SLIP