UNIT TWO SUMMARY NOTES RATIONAL FUNCTIONS :

WHAT IS A RATIONAL FUNCTION ?

}
function
Pc>c) >
Polynomial + pcse) =/ 0

• fcx) =
A Rational Function "
ratio "
que,
is a

must be at least
>
polynomial function t
degree 1 of two
polynomial functions .

•
Rational Functions will have
asymptotes : vertical asymptotes horizontal asymptotes , ,
and oblique asymptotes
✓

not studied this

-
SYMPTOMS year

An asymptote line that the the function approves but

is a
graph of ,
never touches .

•
Vertical Asymptotes
Occurs when the denominator is
equal to zero

Determine the roots of the polynomial function in the denominator to find vertical asymptotes
Ex .

in 3×+5 '
^

I fc
f-
=
>c)
zoe -4
I > or 2×-4=0
- - - - -

?
I
2>c = 4

* x=2 1
This is what
4
a V. A looks like .
a.

It is not a
physical
line but more of an

invisible barrier .

•
Horizontal Asymptotes
If degree of the numerator is less than the degree of the denominator then H.tt is @
y=o
if of the numerator the degree of the denominator
degree is
equal to
^
then Hit is @
lead coefficient
Ex ,
or
I Y =
lead coefficient
4×+5
fc c)
> =
- - - -
l y=z

];
2×-3 I
-
?
equal
- - - - - -

degrees
)
< or are

y=¥=2
o
00
u ✗= 1.5

✓
H A . follows same
idea 4×+1
as V. A bat is in in
fc =

just
>
now horizontal x2 -4 1

I 1
degree in numerator is - - -

1-
- -
"
y :O
loess than denominator C
: ,
'
00
ye 0
I
Ii .z%=z

INTERCEPTS) +
4- INTERLINK)

intercept y intercept
•
•
x -
-

Occiirs When the numerator = 0 Occurs when a- 0

3)c + I 3 >ctl
EX fc c)
>
=
, k - int : 3×+1=0 Ex .
fc > c) =
z , -1
c-
310) I
.
+
32<=-1 l
FCO) =
-

kg
=
-

x = zco) -
l

ooo a -
int @ (5-10) To
y
- int @ (0 1)
,
-
END BEHAVIOURS

•
For vertical asymptotes
1 Pick to the of the
value
right asymptote and sub in in the equation
.
an x -
as x .

2.
Repeat same process
but this time with an a- value to the left of the asymptote .

3. If the y value positive then the end behaviour is

up (increasing ) that side
is
going
-
on

if the then behaviours down ( that side

y value is
negative the end is
going decreasing) on
-
.

42C I 410) -1 ^
!
-

EX .
fcsc) =
2×+2
•
f. ( o) =
zig +2
=

2×+2=0 I

4C 2) l 8 l
%
- - -

,
-

=
22C = -2 •
fc -
2) =
zczy +2
=
-4-+2 I
x= -1 -

,
VI

K=-1

For horizontal
•

asymptotes
1. For end behaviour as x → x sub in 1000 as x .

2. For end behaviour → -1000

as x -
X sub in as x .

3 .
If as x→±x , y → a value
greater than the H.tt , then the end behaviour is above the H.tt ( fac) > H.tl
If as x → IX
, y
→ a value less than the 1-1.1 ,
then the end behaviour is below the H.lt/fc c)> < 1-1. A)

^
42€ 411000) - l I
ÉX fc>c) = •
f ( 1000) =
/ 99 ←
2110003+2
f
=
2×+2
'
.

4=2
- - - - - - - -

→
4. A- = 2 I
4C-10007 - I '

2 / l
•
f- C- 1000)
:
zc-10001+2
=
'

%
✗ = -1

DOMAIN AND RANGE

•
Domain
set of all
possible x -
values for the function
vertical
asymptotes create an exclusion within the domain

Ex .
3×-2
fix =
Gu +3 V. A : 6×+3=0 Domain :{x x # É ,
KEIR }
-
Gx =-3

✗ =
the domain of the function is the set of all
real numbers
excluding
✗ =
.

Range
•

set of all
possible values for the function
y
-

horizontal asymptotes create exclusion within the

range
an

Ex .

32C -2
fees :& É
Range :{ y / }
=
HA
y# ± yer
=
Goers ,

the
range
of the function is the set of all real

numbers
excluding y=É
HOW -10 GRAPH A RATIONAL FUNKO
* specific steps how to determine key characteristics from its equation is
explained above
on
up

1.
Simplify / factor rational function if needed
2. Determine interception
x -
and y
-
intercepted
3. Determine vertical and horizontal
asymptotes)
4. Determine end behaviours for V. A

5. Determine end behaviours for H A .

6. Plot all info on a

graph and connect it all together
7. Check with
graphing technology .

22<+1 2-Ctl
ÉX fc>c) = FC c) =
4C>c. 2) (✗ +2
>
.
4×2-16
if f

_÷
I

[
l
l

I 1
X - int : 2×+1=0
I
a. ÷

an " . >
y.int :
=
-16
- - - - - -
- -
=
←
I ,

7.
Licorice
/ (0-5,0)
I

V. A :
4L>c-2) (2+2)=0 1 I

✗ 1- 2=0 2C -2=0 / /
✗ = -2 2<=2
I

v1 : t !
HA : 2=-2 2=2
.

y=o
21 3) +1-

End behaviours V. A : fc c)
>
=
41.35-16
=
t

2C 1) +I

#F
-

fc>c) =
yup -16
=

2 (1) tl I
f- c c) at
-

=
=
> '
4111 -16

2 (3) + I
fczcj =

41312-16
=
¥ f
24000) +1
End behaviours fc
= 0.0005 above
HA : > c) =
yaooojr.io

21-1000) +1
= -0.0005 below
food =

41-100012-16

Check :
-10W TO SOLVE RATIONAL INEQUALITIES

1. Graph rational function lfcxs)

Graph other side of inequality graph (go.es )
2. on same

3. Determine of values where foes and /or fcx>

glad
> <
range goes

4×+2
5
EX .
2 >e -
l

424-2
za -
i
< 5
when ✗<
É and x >
¥ > 5=II
✗ =
I

•
:÷:÷÷÷
Graphically Ex .
>c-

✗ +1
3
= 0

Graph the function

✗ =3 (3,0)
Look for x -

intercepts •

>
roots are the solution

Algebraically
•

set equation to 0 and determine the set of excluded values

solve for x

X -
l 3, I
xt2_ ✗ =/ -
-

2C +3 ✗ 1- 1
Ex . XI 0
=

:÷=o
7<+1
:÷ -

X -
3--0 or
Get 2) ( act 1) be -1/(20-3)
2C =3 - = 0

(24-3) Get 1) (2+3) (Rtl )

( ✗+2) (act 1) - (x 1) ( act 3)

- = 0

(20+3×+2) (20+2×-3)=0 -

+3kt 2 -
¥ - 2×1--3 = 0

✗ +5=0

2=-5