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Rational & Polynomial Inequalities
Honors Precalculus
Mr. Velazquez
Solving Polynomial Inequalities
A polynomial inequality is any inequality that can be expressed in the forms:
𝑓 𝑥 < 0, 𝑓 𝑥 > 0, 𝑓 𝑥 ≤ 0, or 𝑓 𝑥 ≥ 0
Where 𝑓 𝑥 is a polynomial function.
Solving Polynomial Inequalities
Solving Polynomial Inequalities
Solve and graph the solution set on a real number line: 𝑥2 − 𝑥 ≤ 12
0 2 4 6 8 10 12 14 -14 -12 -10 -8 -6 -4 -2
Then we find the solutions of the quadratic: 𝑥2 − 𝑥 − 12 = 0
(𝑥 − 4)(𝑥 + 3) = 0
𝑥 = 4 𝑥 = −3
−∞, −3 −3, 4 [4, ∞)
First, convert so that the
right side is zero:
𝑥2 − 𝑥 ≤ 12
𝒙𝟐 − 𝒙 − 𝟏𝟐 ≤ 𝟎
These solutions will be the boundaries of our inequality, splitting the
number line into three separate regions. Since the inequality involves
≤ instead of <, we are including both boundary points in our answer:
Solving Polynomial Inequalities
Solve and graph the solution set on a real number line: 𝑥2 − 𝑥 ≤ 12
0 2 4 6 8 10 12 14 -14 -12 -10 -8 -6 -4 -2
For each of these regions, we pick a test point to determine
whether or not to include that region in our solution set.
−∞, −3 −3, 4 [4, ∞)
Test Point: 𝑥 = −4
−4 2 − (−4) ≤ 12
20 ≤ 12
Test Point: 𝑥 = 5
5 2 − (5) ≤ 12
20 ≤ 12
Test Point: 𝑥 = 0
0 2 − (0) ≤ 12
𝟎 ≤ 𝟏𝟐
NOT TRUE NOT TRUE TRUE
Solving Polynomial Inequalities
Solve and graph the solution set on a real number line: 𝑥3 + 𝑥2 − 17𝑥 ≥ −15
0 2 4 6 8 10 12 14 -14 -12 -10 -8 -6 -4 -2
Rearrange:
𝑥3 + 𝑥2 − 17𝑥 + 15 ≥ 0
Factor and find solutions:
𝑥3 + 𝑥2 − 17𝑥 + 15 = 0
𝑥 − 1 𝑥 − 3 𝑥 + 5
𝒙 = 𝟏 𝒙 = 𝟑 𝒙 = −𝟓
−∞, −5 −5, 1 1,3 3, ∞
Solving Polynomial Inequalities
Solve and graph the solution set on a real number line: 𝑥3 + 𝑥2 − 17𝑥 ≥ −15
0 2 4 6 8 10 12 14 -14 -12 -10 -8 -6 -4 -2
−∞, −5
−5, 1
1,3
3, ∞
Test Point: 𝑥 = −6
−6 3 + −6 2 − 17 −6 ≥ −15
−78 ≥ −15
NOT TRUE Test Point: 𝑥 = 0
0 3 + 0 2 − 17 0 ≥ −15
0 ≥ −15
Test Point: 𝑥 = 2
2 3 + 2 2 − 17 2 ≥ −15
−22 ≥ −15
Test Point: 𝑥 = 4
4 3 + 4 2 − 17 4 ≥ −15
12 ≥ −15
NOT TRUE
TRUE TRUE
FIXED
Solving Rational Inequalities
A rational inequality is any inequality that can be expressed in the forms:
𝑓 𝑥 < 0, 𝑓 𝑥 > 0, 𝑓 𝑥 ≤ 0, or 𝑓 𝑥 ≥ 0
Where 𝑓 𝑥 is a rational function.
Solving Rational Inequalities
Find the solution set for the following rational inequality: 3𝑥 + 3
2𝑥 + 4> 0
This inequality is already in the form 𝑓 𝑥 > 0, so we will be finding the
values of 𝑓 that have positive solutions.
x
y
Find the x-intercept by making the
numerator equal to zero:
3𝑥 + 3 = 0
𝑥 = −1
Find where 𝑓 is undefined by making the
denominator equal to zero:
2𝑥 + 4 = 0
𝑥 = −2
Solving Rational Inequalities
Find the solution set for the following rational inequality: 3𝑥 + 3
2𝑥 + 4> 0
You can use test points again here for the intervals (−∞, -2), (-2, 1) and (1,
∞), but we can also see the solution set by looking at the graph
x
y
+ +
The highlighted regions are all
positive, so these are the points
where 𝑓 𝑥 =3𝑥+3
2𝑥+4 is greater than
zero, which satisfies the inequality.
, 2 1,
Solving Rational Inequalities
Find the solution set for the following rational inequality: 5𝑥 − 2
3𝑥 + 1> 0
x-intercept(s):
5𝑥 − 2 = 0
𝑥 =2
5
asymptote(s):
3𝑥 + 1 = 0
𝑥 = −1
3
The function will be positive
in the intervals −∞, −1
3
and 2
5, ∞
−∞, −𝟏
𝟑∪
𝟐
𝟓, ∞
Solving Rational Inequalities
Find the solution set for the following rational inequality: 8𝑥 + 2
4𝑥 − 1< 3
Classwork/Exit Ticket A student is standing on top of a school building which is 80 feet tall. If he throws a ball up in the air with an initial velocity of 32 feet/sec, and the function that represents the height of the ball at time 𝑡 is 𝒉 𝒕 = −𝟏𝟔𝒕𝟐 + 𝟑𝟐𝒕 + 𝟖𝟎 (this includes the starting height), during which time interval will the height of the ball exceed the height of the building? Find the solution set and graph it on a real number line.
Work in pairs or groups of three
to complete the assignment
(each student must hand in
their own work), and turn it in
at the end of class.
Homework:
Rational & Polynomial
Inequalities
Pg. 366, #24–60
(multiples of 4)