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Sect. 9.3 Graphing Rational Functions

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Right of vert. asymp. Example 3. Graph. State domain & range. Vertical asymptote: ... Vert asymp: x 2 = 0. x = 2. 3. x y -1 0. 0 2. 1 6. 3 -4. 4 0. Left of x ... – PowerPoint PPT presentation

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Title: Sect. 9.3 Graphing Rational Functions


1
Sect. 9.3 Graphing Rational Functions
Goal 1 Determine Vertical Asymptotes and
Point of Discontinuity Goal 2 Graph Rational
Functions
2
Rational Function
A function of the form
where p(x) and q(x) are polynomial functions and
q(x) ? 0.
Examples
3
Graphs of Rational Functions may have breaks in
Continuity
Breaks in Continuity can appear as
1. Vertical Asympotes
2. Point Discontinuity
A hole in the graph.
4
Vertical Asymptote
If a rational expression is written in simplest
form and the function is undefined for x a,
then x a is a vertical asymptote.
Example
x - 2 is vertical asymptote
5
Point Discontinuity
If the original function is undefined for x a
but the rational expression of the function in
simplest form is defined for x a, then there is
a hole in the graph at x a.
Example
Point of Discontiniuty as x -2
6
Determine the equations of any vertical
asymptotes and the values of x for any holes for
the following function.
x - 3 is a vertical aymptote
x - 2 is a hole in the graph
7
Graph
8
Graph State the domain range.
Vertical Asymptote x 1 Horizontal Asymptote
y 2
x y -5 1.5 -2 1 0 -1 4 3
Left of vert. asymp.
Right of vert. asymp.
Domain all real s except 1. Range all real
s except 2.
9
GraphState domain range.
Vertical asymptote 3x3 0 (set denominator
0) 3x -3 x -1
x y -3 .83 -2 1.33 0 -.67 2
0
Domain All real s except -1. Range All real
s except 1/3.
10
SOLUTION
The numerator has no zeros, so there is no
x-intercept.
The denominator has no real zeros, so there is no
vertical asymptote.
The bell-shaped graph passes through (3, 0.4),
( 1, 2), (0, 4), (1,2), and (3, 0.4). The
domain is all real numbers the range is 0 lt y
4.
11
SOLUTION
The numerator has 0 as its only zero,so the
graph has one x-intercept at (0, 0).
The denominator can be factored as(x 2)(x
2), so the denominator haszeros at 2 and 2.
This implies vertical asymptotes at x 2 and
x 2.
12
To draw the graph, plot points between and beyond
vertical asymptotes.
13
SOLUTION
The numerator can be factored as ( x 3) and ( x
1) the x-intercepts are 3 and 1.
The only zero of the denominator is 4, sothe
only vertical asymptote is x 4 .
14
To draw the graph, plot points to the left and
right of the vertical asymptote.
15
Graph. State domain range.
4. x y -2 -.4 -1 -.5
0 0 1 .5 2 .4
  • x-intercepts x 0
  • vert. asymp. x2 1 0
  • x2 -1
  • No vertical
  • asymptote

(No real solutions)
16
Domain all real numbers Range
17
Graph, then state the domain and range.
  • x-intercepts
  • 3x2 0
  • x2 0
  • x 0
  • Vert asymp
  • x2- 4 0
  • x2 4
  • x 2 x -2

3. x y 4 4 3 5.4 1
-1 0 0 -1 -1 -3
5.4 -4 4
On right of x2 asymp.
Between the 2 asymp.
On left of x-2 asymp.
18
Domain all real s except -2 2 Range all
real s except 3
19
Graph, then state the domain range.
  • x-intercepts
  • x2 - 3x 4 0
  • (x - 4)(x 1) 0
  • x 4 0 x 1 0
  • x 4 x -1
  • Vert asymp
  • x 2 0
  • x 2

3. x y -1 0 0
2 1 6 3 -4 4
0
Left of x2 asymp.
Right of x2 asymp.
20
Domain all real s except 2 Range all real
s
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