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Math-3Lesson 4-5
Polynomial and Rational Inequalities
What would you call this?
x > 3 and x < 5
Compound inequality: 2 inequalities
joined together by either “and” or “or”.
1 2 3 4 5
What part is x > 3 and x < 5 ??
x ≤ 3 or x > 5
What would it look like on the number line?
1 2 3 4 5
3 and 5 divide the solution set (of numbers)
from those numbers not belonging to the
solution set.
The following are some inequalities we will solve.
)2)(3(0 xx
1660 2 xx
)2)(1)(1(0 xxx
xxx 540 23
4
3
1. Find the “real” zeroes of the polynomial equation.
2. Divide the x-axis into intervals using the
zeroes of the polynomial.
120 2 xx
120 2 xx 0 = (x – 4)(x + 3)
4,3x
3. Determine the sign (+/-) of the polynomial
for each interval of x-values. 122 xxy3. Determine the interval of x-values that
make the inequality true. )4,3(
-2 and 6 divide the solution from the “non-solution.”
Solve the equation:
1240 2 xx 1240 2 xx
0 = (x – 6)(x + 2)
-2 6
The number line has been divided into 3 intervals:
),6()6,2()2,(
The number line has been divided into 3 intervals:
Pick any number in the left-most interval and
substitute it in for x in the intercept form equation.
1240 2 xx-2 6
),6()6,2()2,(
)6)(2( xxythe number inside 1st parentheses will be
negative for any number in the left-most interval.
)6)(( xythe number inside 2nd parentheses will be
negative for any number in the left-most interval.
))(( y A negative times a negative is positive
)(y The output is positive for this interval.
+
1240 2 xx-2 6
),6()6,2()2,(
)6)(2( xxy
sign of number inside 1st parentheses will be
positive for any number in the middle interval.
)6)(( xy
Middle interval of ‘x’:
))(( y A positive times a negative is negative
)(y The output is negative for this interval.
+
the sign of the number inside 2nd parentheses will
be negative for any number in the middle interval.
-
1240 2 xx-2 6
),6()6,2()2,(
)6)(2( xxy
sign of number inside 1st parentheses will be
positive for any number in the right side interval.
)6)(( xy
Right side interval of ‘x’:
))(( y A positive times a positive is positive
)(y The output is positive for this interval.
+
the sign of the number inside 2nd parentheses will
be positive for any number in the right side interval.
- +
1240 2 xx
-2 6
),6()6,2()2,(
Which interval(s) make the inequality true?
What is the solution to the inequality?
+ - +
),6()2,(
1242 xxy
-2 6
Graph the general shape of the equation.
+-
+
Positive lead coefficient, even degree
Up on left and right
No even multiplicities
What do you notice about
the graph and the signs for
each interval of ‘x’?
Same: you could
solve the inequality by
looking at the sign of ‘y’
from the graph!!!
1240 2 xx
Your turn: solve the inequality
xx 45 2
When solving quadratic equations, we first
rearranged the equation to be in standard form.
540 2 xx
540 2 xx
-1 5Solution occurs in the intervals
where the output is “+” )1,( x For
540 2 xx )1)(5(0 xx
),5()5,1()1,(
))(( y
+ -
)5,1( x For
))(( y
)1)(5( xxy
),5( x For
))(( y
+
Solving Single Variable Rational Inequalities
We will solve inequalities similar to the following:
)1)(2(
)2)(6(30
xx
xx
166
1070
2
2
xx
xx
16
630
2
x
x
)4)(4(
)2(3
xx
xy
24 4
1) Passes thru at
x = 2
2) VA at x = -4, 4
)())((
)(3
),4()4,2()2,4()4,(
)())((
)(3
)(
))((
)(3
)(
))((
)(3
),4()2,4( x
23
361230
2
2
xx
xx
Vertical asymptote:
X-intercept:
x = -2, -1
6,2 x
)1)(2(
)124(3 2
xx
xxy
)1)(2(
)2)(6(3
xx
xxy
22 1
6
)())((
))((3
),2()2,1()1,2()2,6()6,( )(
))((
))((3
)())((
))((3
)())((
))((3
)())((
))((3
)2,1()2,6( x
12
1660
2
2
xx
xx
)4)(3(
)2)(8(
xx
xxy
83 4
2
Your turn: analyze the graph using ‘sign changes’.
12
1662
2
xx
xxy
Vertical asymptote:
X-intercept:
x = -3, 4
8,2x
))((
))(()5.2(
f
)4)(3(
)2)(8(
xx
xxy
83 4
2
1) Passes thru at x = -2, 8
2) For x < -3, y is either
“+” or “-”, analyze at
some number smaller
than -3
))((
))((
y ))((
))(()0(
f
Your turn: analyze the graph using ‘sign changes’.
3) For -3 < x -2, y is either “+”
or “-”, analyze at -2.5
4) For -2 < x 4, y is either “+”
or “-”, analyze at 0
44
281022
2
xx
xxy
2. Vertical asymptote:
1. X-intercept:
x = -2
7,2x
)()(
)(2)5(
f
)2)(2(
)2)(7(2
xx
xxy
72
2at x hole
Passes thru (no kiss)
at x = -2, 7
4. For x < -2, y is either
“+” or “-”, analyze at
some number smaller
than -2 (-3 is nice)
)(
)(2)3(
f
)()(
)(2)10(
f
Your turn: analyze the graph using ‘sign changes’.
6. For 2 < x 7, y is either “+” or
“-”, analyze at 5
7. For x > 7, y is either “+” or
“-”, analyze at 10
)2(
)7(2
x
x
3. Hole: x = -2
2
5. For -2 < x 2, y is either “+”
or “-”, analyze at 0)(
)(
)(2)0(
f
)3(f
16
271832
2
x
xxy
Vertical asymptote:
X-intercept:
x = -4, 4
3x)(
))((
))((3)5.3(
f
)4)(4(
)3)(3(3
xx
xxy
4
3
1) Kisses at x = -3
2) For x < -4, y is either
“+” or “-”, analyze at
some number smaller
than -4 (at -5)
))((
))((3)5(
f
)())((
))((3)5(
f
Your turn: analyze the graph using ‘sign changes’.
3) For -4 < x -3, y is either
“+” or “-”, analyze at -3.5
4) For x > 4, y is either
“+” or “-”, analyze at 5
Hole:
none
4
)()5( f
16
271832
2
x
xxy
)4)(4(
)3)(3(3
xx
xxy
4
3
Your turn: analyze the graph using ‘sign changes’.
4
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