# Polynomial & Rational Inequalities

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"Polynomial & Rational Inequalities"

```					     Analyzing Polynomial & Rational
Functions & Their Graphs
Steps in Analysis of Graphs of Poly-Rat Functions
1)    Examine graph for the domain with attention to
holes. (If f = p/q, “holes” are where q(x) = 0.) Here
there will be vertical asymptotes.
2)    Find any root(s) where f(x) = 0 or, if f = p/q,
p(x) = 0. Note where f(x) = 0, p(x) = 0, or q(x) = 0,
they may be factored.
3)    Record behavior specific to intervals of roots/holes.
4)    Determine any symmetry properties and any
horizontal or oblique asymptotes.
Polynomial & Rational
Inequalities
Steps to Solution & Graph of Poly-Rat (In)equalities
1) Write as form: f(x) > 0, f(x) > 0, f(x) < 0, f(x) < 0,
or f(x) = 0, with single quotient if f is rational.
2) Find any root(s) of f(x) = 0 &, if f = p/q, find
“holes” where q(x) = 0. Factor f , p, q as possible.
3) Separate real number line into intervals per above.
4) For an x = xi in each interval find f(xi). Note:
sign[f(xi)] = sign[f(x)] for xi in interval i. Use this
to sketch graph. Also, if f inequality was > or < ,
include in solution set roots of f from 2) above.
Poly-Rat Inequalities Example
Solve & graph:
(x + 3)     > (x + 3) .
(x2 - 2x + 1)     (x – 1)
Step 1:
(x + 3) –   (x + 3)(x – 1)   > 0.
(x – 1)2    (x – 1)(x – 1)
(x + 3)(2 – x) > 0
___ (x – 1)2
or f(x) = p(x)/q(x) > 0 with
p(x) = (x + 3)(2 – x) and q(x) = (x – 1)2.
Poly-Rat Inequalities Example cont’d
Solve & graph:
(x + 3)(2 – x) > 0
___                    (x – 1)2
Step 2: Note roots of f(x) = roots of p(x).
They are at x = –3 and at x = 2.
The point x = 1 is a zero of q(x) so there f(x) is
undefined and x = 1 is a “hole” or not in the domain.
Step 3:
The intervals: (-, -3]; [-3, 1);     (1, 2];    [2, ).
f(x) values: f(-4)= -6/25, f(0)= 6, f(3/2)= 9, f(3)= -3/2
Poly-Rat Inequalities Example cont’d
Solve & graph:
(x + 3)(2 – x) > 0
___                 (x – 1)2
Step 2 & 3 Data Summery:
x-intercepts: at x = –3 and at x = 2.
y-intercept: at y = f(0) = 6.
f(-x) = (-x + 3)(2 + x)     f(-x) _
_          (-x – 1)2
No symmetry.
Poly-Rat Inequalities Example cont’d
Solve & graph:
(x + 3)(2 – x) > 0
___                 (x – 1)2
Step 2 & 3 Data Summery cont’d:
Vertical asymptote: at x = 1.
Hole: x = 1

Horizontal asymptote: at y = -1
Intervals:
- < x < -3, -3 < x < 1, 1 < x < 2, 2 < x < .
Poly-Rat Inequalities Example cont’d
f(x) = (x + 3)(2 – x)/(x – 1)2.
   x  3  3  x  11  x  2 2  x  
-4               0           3/2          3
f(-4) = -6/25    f(0) = 6      f(3/2) = 9   f(3) = -3/2

Below            Above         Above      Below
x-axis           x -axis       x -axis    x-axis
(-4, -6/25)        (0, 6)        (3/2, 9)   (3, -3/2)
Poly-Rat Inequalities Example cont’d
Solve & graph:
(x + 3)(2 – x) > 0
___                  (x – 1)2
Step 4 Graphs: A) Solution set as intervals on the
number line –
(-, -3];        [-3, 1); (1, 2]; [2, ).
neg      0           pos  pos 0         neg

[           )( ]
-3 -2 -1 0 1 2
Poly-Rat Inequalities Example cont’d
Solve & graph:
(x + 3)(2 – x) > 0
___                     (x – 1)2
Step 4 Graphs: B) Solution set in graph sketch of f(x)
versus x. First plot known points. Then sketch.
Do not forget in sketching to include information
about asymptotes. In this case, since x = 1 is a zero of
multiplicity 2 in q(x), there is a vertical asymptote at
x = 1.
Also, since both p(x) and q(x) are of 2nd degree, there
is a horizontal asymptote at y = – 1/1 as |x| increases.
Poly-Rat Inequalities Example cont’d
Solve & graph:
(x + 3)(2 – x) > 0
___                        (x – 1)2
9-

6-

3-

        
-5 -4 -3 -2 -1 0  2 3 4 5
1
Hole & Asymptotes: (1, (3/2, f(x) graph in
Intercepts: (-3, 0), of (0, 6), 1, y (2,
Sketch
Sketchf(x) = (3, -3/2)
of 9), = -2
Test values: (-4, -6/25), 0), x > 0 graph 0) red

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