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Warm-Up Exercises
1. Solve x2 – 2x – 24 = 0.
ANSWER
2. Solve
ANSWER
–4
–8 < 3x –5 < 7.
–1 < x < 4
Warm-Up Exercises
3. A point on a graph moves so that the function
y = 2x2 – 3x – 20 describes how the vertical
position y is related to the horizontal position x.
What is the vertical position when the horizontal
position is –3?
ANSWER
7
Graph a quadratic inequality
Warm-Up1Exercises
EXAMPLE
Graph y > x2 + 3x – 4.
SOLUTION
STEP 1
Graph y = x2 + 3x – 4. Because the
inequality symbol is >, make the
parabola dashed.
STEP 2
Test a point inside the parabola,
such as (0, 0).
y > x2 + 3x – 4
?
0 > 02 + 3(0) – 4
0>–4
Graph a quadratic inequality
Warm-Up1Exercises
EXAMPLE
So, (0, 0) is a solution of the inequality.
STEP 3
Shade the region inside the parabola.
Use a quadratic inequality in real life
Warm-Up2Exercises
EXAMPLE
Rappelling
A manila rope used for rappelling down a cliff can
safely support a weight W (in pounds) provided
W ≤ 1480d2
where d is the rope’s diameter (in inches).
Graph the inequality.
SOLUTION
Graph W = 1480d2 for nonnegative values of d.
Because the inequality symbol is ≤, make the
parabola solid. Test a point inside the parabola, such
as (1, 2000).
Use a quadratic inequality in real life
Warm-Up2Exercises
EXAMPLE
W ≤ 1480d2
2000 ≤ 1480(1)2
2000 ≤ 1480
Because (1, 2000) is not a solution, shade the region
below the parabola.
Graph a system of quadratic inequalities
Warm-Up3Exercises
EXAMPLE
Graph the system of quadratic inequalities.
Inequality 1
y < – x2 + 4
y > x2 – 2x – 3 Inequality 2
SOLUTION
STEP 1
Graph y ≤ – x2 + 4. The graph is the
red region inside and including the
parabola y = – x2 + 4.
Graph a system of quadratic inequalities
Warm-Up3Exercises
EXAMPLE
STEP 2
Graph y > x2– 2x – 3. The graph is the blue region inside
(but not including) the parabola y = x2 –2x – 3.
STEP 3
Identify the purple region where
the two graphs overlap. This region
is the graph of the system.
Warm-Up
YOU
TRY. . Exercises
.
for Examples 1, 2, and 3
Graph the inequality.
1. y > x2 + 2x – 8
STEP 1
Graph y = x2 + 2x – 8. Because the
inequality symbol is >, make the
parabola dashed.
STEP 2
Test a point inside the parabola,
such as (0, 0).
y > x2 + 2x – 8
? 2
0 > 0 + 2(0) – 8
0>–4
Warm-Up
YOU
TRY. . Exercises
.
for Examples 1, 2, and 3
So, (0, 0) is a solution of the inequality.
STEP 3
Shade the region inside the parabola.
Warm-Up
YOU
TRY. . Exercises
.
for Examples 1, 2, and 3
Graph the inequality.
2. y < 2x2 – 3x + 1
SOLUTION
STEP 1
Graph y = 2x2 – 3x + 1. Because the
inequality symbol is <, make the
parabola dashed.
STEP 2
Test a point inside the parabola,
such as (0, 0).
y < 2x2 – 3x + 1
? 2
0 < 0 – 3(0) + 1
0<1
Warm-Up
YOU
TRY. . Exercises
.
for Examples 1, 2, and 3
So, (0, 0) is a solution of the inequality.
STEP 3
Shade the region inside the parabola.
Warm-Up
YOU
TRY. . Exercises
.
for Examples 1, 2, and 3
Graph the inequality.
3. y < – x2 + 4x + 2
SOLUTION
STEP 1
Graph y = – x2 + 4x + 2. Because the
inequality symbol is <, make the
parabola dashed.
STEP 2
Test a point inside the parabola,
such as (0, 0).
y < – x2 + 4x + 2
? 2
0 < 0 + 4(0) + 2
0<2
Warm-Up
YOU
TRY. . Exercises
.
for Examples 1, 2, and 3
So, (0, 0) is a solution of the inequality.
STEP 3
Shade the region inside the parabola.
Warm-Up
YOU
TRY. . Exercises
.
for Examples 1, 2, and 3
4. Graph the system of inequalities consisting of
y ≥ x2 and y < 2x2 + 5.
SOLUTION
STEP 1
Graph y –> x2.
STEP 2
Graph y < 2x2 + 5.
STEP 3
Identify the shaded region where the two graphs
overlap. This region is the graph of the system.
Solve a quadratic inequality using a table
Warm-Up4Exercises
EXAMPLE
Solve x2 + x ≤ 6 using a table.
SOLUTION
Rewrite the inequality as x2 + x – 6 ≤ 0.
Then make a table of values.
Notice that x2 + x –6 ≤ 0 when the values of x are
between –3 and 2, inclusive.
ANSWER
The solution of the inequality is –3 ≤ x ≤ 2.
Solve a quadratic inequality by graphing
Warm-Up5Exercises
EXAMPLE
Solve 2x2 + x – 4 ≥ 0 by graphing.
SOLUTION
The solution consists of the x-values for which the
graph of y = 2x2 + x – 4 lies on or above the x-axis. Find
the graph’s x-intercepts by letting y = 0 and using the
quadratic formula to solve for x.
0 = 2x2 + x – 4
x = – 1+ 12– 4(2)(– 4)
2(2)
– 1+ 33
x= 4
x 1.19 or x
–1.69
Solve a quadratic inequality by graphing
Warm-Up5Exercises
EXAMPLE
Sketch a parabola that opens
up and has 1.19 and –1.69 as
x-intercepts. The graph lies
on or above the x-axis to the
left of (and including)
x = – 1.69 and to the right of
(and including) x = 1.19.
ANSWER
The solution of the inequality is approximately
x ≤ – 1.69 or x ≥ 1.19.
Warm-Up
YOU
TRY. . Exercises
.
5.
for Examples 4 and 5
Solve the inequality 2x2 + 2x ≤ 3 using a table and
using a graph.
SOLUTION
Rewrite the inequality as 2x2 + 2x – 3 ≤ 0.
Then make a table of values.
x
-3
-2
-1.8
-1.5
-1
0
0.5
0.8
0.9
22 + 2x – 3
9
1
-0.1
-1.5
-3
-3
-1.5
-0.1
0.42
ANSWER
The solution of the inequality is –1.8 ≤ x ≤ 0.82.
Warm-Up6Exercises
EXAMPLE
Use a quadratic inequality as a model
Robotics
The number T of teams that
have participated in a robotbuilding competition for high
school students can be
modeled by
T(x) = 7.51x2 –16.4x + 35.0, 0 ≤ x ≤ 9
Where x is the number of years since 1992.
For what years was the number of teams
greater than 100?
Warm-Up6Exercises
EXAMPLE
Use a quadratic inequality as a model
SOLUTION
You want to find the values of x
for which:
T(x) > 100
7.51x2 – 16.4x + 35.0 > 100
7.51x2 – 16.4x – 65 > 0
Graph y = 7.51x2 – 16.4x – 65 on the domain 0 ≤ x ≤ 9.
The graph’s x-intercept is about 4.2. The graph lies
above the x-axis when 4.2 < x ≤ 9.
ANSWER
There were more than 100 teams participating in
the years 1997–2001.
Warm-Up7Exercises
EXAMPLE
Solve a quadratic inequality algebraically
Solve x2 – 2x > 15 algebraically.
SOLUTION
First, write and solve the equation obtained by
replacing > with = .
x2 – 2x = 15
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = – 3 or x = 5
Write equation that corresponds
to original inequality.
Write in standard form.
Factor.
Zero product property
Warm-Up7Exercises
EXAMPLE
Solve a quadratic inequality algebraically
The numbers – 3 and 5 are the critical x-values of the
inequality x2 – 2x > 15. Plot – 3 and 5 on a number line,
using open dots because the values do not satisfy the
inequality. The critical x-values partition the number
line into three intervals. Test an x-value in each
interval to see if it satisfies the inequality.
Test x = – 4:
(– 4)2 –2(– 4) = 24 >15
Test x = 1:
12 –2(1) 5 –1 >15
ANSWER
The solution is x < – 3 or x > 5.
Test x = 6:
62 –2(6) = 24 >15
Warm-Up
YOU
TRY. . Exercises
.
6.
for Examples 6 and 7
Robotics
Use the information in Example 6 to determine in
what years at least 200 teams participated in the
robot-building competition.
SOLUTION
You want to find the values of x for which:
T(x) > 200
7.51x2 – 16.4x + 35.0 > 200
7.51x2 – 16.4x – 165 > 0
Warm-Up
YOU
TRY. . Exercises
.
for Examples 6 and 7
Graph y = 7.51x2 – 16.4x – 165 on the domain 0 ≤ x ≤ 9.
ANSWER
There were more than 200 teams participating in
the years 1998 – 2001.
Warm-Up
YOU
TRY. . Exercises
.
7.
for Examples 6 and 7
Solve the inequality 2x2 – 7x = 4 algebraically.
SOLUTION
First, write and solve the equation obtained by
replacing > with 5.
2x2 – 7x = 4
2x2 – 7x – 4 = 0
(2x + 1)(x – 4) = 0
x = – 0.5 or x = 4
Write equation that corresponds
to original inequality.
Write in standard form.
Factor.
Zero product property
Warm-Up
YOU
TRY. . Exercises
.
for Examples 6 and 7
The numbers 4 and – 0.5 are the critical x-values of the
inequality 2x2 – 7x > 4 . Plot 4 and – 0.5 on a number
line, using open dots because the values do not
satisfy the inequality. The critical x-values partition the
number line into three intervals. Test an x-value in
each interval to see if it satisfies the inequality.
–7 –6 –5 –4 –3 –2 –1
Test x = – 3:
0
1
2
Test x = 2:
2 (– 3)2 – 7 (– 3) > 4  2 (2)2 – 7 (2) > 4
ANSWER
The solution is x < – 0.5 or x > 4.
3
4
5
6
7
Test x = 5:
2 (5)2 – 7 (3) > 4 
Warm-Up
Exercises
KEEP
GOING
1.
Solve 3x2 + 2x – 1 < 0 by graphing.
ANSWER
2.
– 1 < x < 13
Graph y  2x2 – 4 and y < – x2 + 3 as a system of
inequalities.
ANSWER
Warm-Up
Exercises
KEEP
GOING
Solve the inequalities.
3.
x2 + 4x  5
ANSWER
4.
–5x1
2x2 + x – 15 > 0
ANSWER
x < – 3 or x > 52