Notes
11.3
POINT OF DISCONTINUITY
x-coordinate
A point of discontinuity is the __________________
of a point where the graph of a function f(x) is
________
______________________
continuous
not
Points of discontinuity have already been introduced
in this unit as asymptotes in the reciprocal function
graph . In this section, we will look at many rational
functions. We will identify additional points of
discontinuity called “holes” in graphs.
REMOVABLE AND NON-REMOVABLE
DISCONTINUITY
Removable
Non-removable
 ________
Hole in the graph
 ____________
Asymptote in the graph
 Could be ________________
continuous
 There is no way to if
if _____________
the function
redefine
at that point
___________
redefine the function at that
point to make it continuous
 The hole occurs in the equation
where there is a common factor
in the numerator and
denominator, thus the
______________
ORIGINAL function is
______________
undefined at a specific
x-value
By the way,
continuous graphs
have no jumps,
breaks, holes or
asymptotes!!
LETS EXPLORE HOW
TO
FIND POINTS OF
DISCONTINUITY
FIND POINT OF DISCONTINUITY : ASYMPTOTES
Horizontal Asymptote:
If Exponent is:
y=___
BOB0
Bigger on bottom…0
BOTN
Bigger on top…none
EATS
DC
Exponents are the same
Asymptote is
Y=0
No H.A.
Divide Coefficients
of leading term
Vertical Asymptote: Simplify Function and set denominator
equal to zero and solve for x
x=___
Ex.
f ( x) 
x 1
2
x4
H. Asymptote: EATS DC
y=3
Divide coefficients and get
1 for the fraction but then
add the constant of 2
V. Asymptote:
Set denominator equal to 0
X = -4
Pick 2 x-values to left
and right of V.A.
X
Y
-6
4.5
-5
6
-4
und
-3
0
-2
1.5
This is non-removable discontinuity bc you can’t
redefine the fcn and make the graph continuous
FIND POINT OF DISCONTINUITY : HOLE(S)
1. Look at the given equation.
2. Factor the numerator and denominator
3. If there is a common factor, set that
factor equal to zero and solve for x.
4. This x-value is where there will be a hole
in the graph.
EX:
( x  2 x  3) ( x  3)( x  1)

Graph y 
( x  3)
( x  3)
2
 x 1
BOTN so no H.A.
x+3 either causes a
V.A. or a hole…
Because x+3
factors out, there
is a hole in the
graph at x = – 3
This is removable discontinuity bc you can redefine
the fcn at f(-3)=-4 and make the graph continuous
Ex
y
x  x  12 ( x  4)( x  3)

2
x 4
( x  2)( x  2)
2
H. Asymptote:
EATS DC
y=1
Divide leading coefficients
and get 1 for the fraction
Hole:
X
Y
-4
0
-1.2
-3
-1
0
1
4
3
3
10/3
0
4
2/3
Factor to see if anything
factors out.
Nothing factors out so there is
no hole in the graph.
V. Asymptote:
Set denominator factors equal to 0
x = 2 and x = – 2
Pick x-values to left
and right of V.A..
This is non-removable discontinuity bc you can’t
redefine the fcn and make the graph continuous
THE FUNCTION BELOW GIVES THE
CONCENTRATION OF THE SALINE SOLUTION
AFTER ADDING X ML OF THE 0.5% SOLUTION TO
100 ML OF THE 2% SOLUTION.
(100)(0.02)  x(0.005)
y
100  x
How many mL of the 0.5% solution must you add for the
combined solution to have a concentration of 0.9%?
275mL of 0.5% solution
0.009(100  x)  2  .005 x
.9  .009x  2  .005x
.004x  1.1
x  275
to get a .9% solution
You can also check this on the
graphing calculator. Type the right
side of equation into y1 and input
0.009 into y2. Use the intersection
function (2nd calc 5) to solve for x.
Opt
f ( x) 
Oblique Asymptote
(not tested)
x  2x  3
x3
2
H. Asymptote: BOTN
Bigger by one degree so
there is an oblique asymptote
Divide using synthetic division
–3
1
y=x-5 R12
y = x – 5 is the oblique
asymptote
Graph on the TI84 to see
what it looks like.
1
–2
–3
–3
15
–5
12
Vocab clarification : Match the following then discuss
11.3 Vocabulary Support Attributes of Rational Functions Concept List
A.
continuous
B. discontinuous
D. horizontal asymptote
E. non-removable discontinuity
G. rational function
H. removable discontinuity
C. factors
F. point of discontinuity
I. vertical asymptote
Choose the concept from the list above that best represents the item in each box.
1. the line that a graph
approaches as y
increases in absolute
value
4. This type of graph has
no jumps, breaks, or
holes.
7. This type of discontinuity
appears as a vertical
asymptote on the graph.
2. In the denominator, these
reveal the points of
discontinuity.
3. This type of
discontinuity appears as
a hole in the graph.
5. a function that you
6. a graph that has a one-
can write in the form
P ( x)
f ( x) =
where
Q( x)
P(x) and Q(x) are
polynomial functions
8. The graph of f (x) is
not continuous at this
point.
point hole or a vertical
asymptote
9. the line that a graph
approaches as x
increases in absolute
value
HOMEWORK:
WS 11.3