Graphing Quadratic Inequalities
4.9
EXAMPLE 1 Graph a quadratic inequality
Graph y > x2 + 3x – 4.
SOLUTION
STEP 1
Graph y = x2 + 3x – 4. Because the inequality symbol is >, make the parabola dashed.
Test a point inside the parabola, such as (0, 0).
STEP 2
y > x2 + 3x – 4
0 > 02 + 3(0) – 4?
0 > – 4
EXAMPLE 1 Graph a quadratic inequality
So, (0, 0) is a solution of the inequality.
STEP 3
Shade the region inside the parabola.
EXAMPLE 2 Use a quadratic inequality in real life
A manila rope used for rappelling down a cliff can safely support a weight W (in pounds) provided
Rappelling
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).
EXAMPLE 2 Use a quadratic inequality in real life
W ≤ 1480d2
2000 ≤ 1480(1)2
2000 ≤ 1480
Because (1, 2000) is not a solution, shade the region below the parabola.
EXAMPLE 3 Graph a system of quadratic inequalities
Graph the system of quadratic inequalities.
y < – x2 + 4 Inequality 1y > 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.
EXAMPLE 3 Graph a system of quadratic inequalities
STEP 2
Graph y > x2– 2x – 3. The graph is the blue region inside (but not including) the parabola y = x2 –2x – 3.
Identify the purple region where the two graphs overlap. This region is the graph of the system.
STEP 3
GUIDED PRACTICE 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.
Test a point inside the parabola, such as (0, 0).
STEP 2
y > x2 + 2x – 8
0 > 02 + 2(0) – 8?
0 > – 4
GUIDED PRACTICE for Examples 1, 2, and 3
So, (0, 0) is a solution of the inequality.
STEP 3
Shade the region inside the parabola.
GUIDED PRACTICE for Examples 1, 2, and 3
Graph the inequality.
y < 2x2 – 3x + 12.
SOLUTION
STEP 1
Graph y = 2x2 – 3x + 1. Because the inequality symbol is <, make the parabola dashed.
Test a point inside the parabola, such as (0, 0).
STEP 2
y < 2x2 – 3x + 1
0 < 02 – 3(0) + 1?
0 < 1
GUIDED PRACTICE for Examples 1, 2, and 3
So, (0, 0) is a solution of the inequality.
STEP 3
Shade the region inside the parabola.
GUIDED PRACTICE for Examples 1, 2, and 3
Graph the inequality.
y < – x2 + 4x + 23. SOLUTION
STEP 1
Graph y = – x2 + 4x + 2. Because the inequality symbol is <, make the parabola dashed.
Test a point inside the parabola, such as (0, 0).
STEP 2
y < – x2 + 4x + 2
0 < 02 + 4(0) + 2?
0 < 2
GUIDED PRACTICE for Examples 1, 2, and 3
So, (0, 0) is a solution of the inequality.
STEP 3
Shade the region inside the parabola.
GUIDED PRACTICE for Examples 1, 2, and 3
4. Graph the system of inequalities consisting of y ≥ x2 and y < 2x2 + 5.
SOLUTION
STEP 1
STEP 2
Graph y < 2x2 + 5.
Graph y > x2.–
Identify the shaded region where the two graphs overlap. This region is the graph of the system.
STEP 3
EXAMPLE 4 Solve a quadratic inequality using a table
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.
The solution of the inequality is –3 ≤ x ≤ 2.
ANSWER
EXAMPLE 5 Solve a quadratic inequality by graphing
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)
x =– 1+ 33
4
x 1.19 or x –1.69
EXAMPLE 5 Solve a quadratic inequality by graphing
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.
The solution of the inequality is approximately x ≤ – 1.69 or x ≥ 1.19.
ANSWER
GUIDED PRACTICE for Examples 4 and 5
Solve the inequality 2x2 + 2x ≤ 3 using a table and using a graph. 5.
SOLUTION
Rewrite the inequality as 2x2 + 2x – 3 ≤ 0. Then make a table of values.
The solution of the inequality is –1.8 ≤ x ≤ 0.82.
ANSWER
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
EXAMPLE 6 Use a quadratic inequality as a model
The number T of teams that have participated in a robot-building competition for high school students can be modeled by
Robotics
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?
EXAMPLE 6 Use a quadratic inequality as a model
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.
There were more than 100 teams participating in the years 1997–2001.
ANSWER
SOLUTION
You want to find the values of x for which:
EXAMPLE 7 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
EXAMPLE 7 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: Test x = 1:
12 –2(1) 5 –1 >15
Test x = 6:
The solution is x < – 3 or x > 5.
ANSWER
(– 4)2 –2(– 4) = 24 >15 62 –2(6) = 24 >15
GUIDED PRACTICE for Examples 6 and 7
6. 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
GUIDED PRACTICE for Examples 6 and 7
Graph y = 7.51x2 – 16.4x – 165 on the domain 0 ≤ x ≤ 9.
There were more than 200 teams participating in the years 1998 – 2001.
ANSWER
GUIDED PRACTICE for Examples 6 and 7
7. Solve the inequality 2x2 – 7x = 4 algebraically.
SOLUTION
First, write and solve the equation obtained by replacing > with 5.
Write equation that corresponds to original inequality.
Write in standard form.
Factor.
Zero product property
2x2 – 7x = 4
2x2 – 7x – 4 = 0
(2x + 1)(x – 4) = 0
x = – 0.5 or x = 4
GUIDED PRACTICE 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.
Test x = – 3: Test x = 2: Test x = 5:
– 3
– 4
– 2
– 1
0
1 2
3
4
5
6
7
– 5
– 6
– 7
2 (– 3)2 – 7 (– 3) > 4 2 (5)2 – 7 (3) > 4 2 (2)2 – 7 (2) > 4
The solution is x < – 0.5 or x > 4.
ANSWER