1. Objectives Solve quadratic inequalities by using tables and graphs.
Solve quadratic inequalities by using algebra.

2. Many business profits can be modeled by quadratic functions.
To ensure that the profit is above a certain level, financial planners may need to graph and solve quadratic inequalities.

3. To graph Quadratic Inequalities
y < ax2 + bx + c y > ax2 + bx + c use a dashed line
y ≤ ax2 + bx + c y ≥ ax2 + bx + c use a solid line
If the parabola opens up:
> or shade inside
< or shade outside
If the parabola opens down:
> or shade outside
< or shade inside

4. Example 1: Graphing Quadratic Inequalities in Two Variables
Graph y ≥ x2 – 7x + 10.
Step 1 Graph the boundary of the related parabola
vertex
(3.5, -2.25)
roots
(2, 0) and (5, 0)
y-intercept
(0, 10)
reflection of y-intercept
(7, 10)

5. Example 1 Continued
Step 2 Shade above the parabola because the solution consists of y-values greater than those on the parabola for corresponding x-values.

6. Example 2
Graph each inequality.
y < –3x2 – 6x – 7
Step 1 Graph the boundary
vertex (-1, -4)
no roots
y-intercept (0, -7)
reflect of y-int (-2, -7)
with a dashed curve.

7. Example 2
Step 2 Shade below the parabola because the solution consists of y-values less than those on the parabola for corresponding x-values.

8. Quadratic inequalities in one variable, such as ax2 + bx + c > 0 (a ≠ 0), have solutions in one variable that are graphed on a number line.
For and statements, both of the conditions must be true. For or statements, at least one of the conditions must be true.
Reading Math

9. you can solve quadratic inequalities algebraically.
By finding the critical values,
you can solve quadratic
inequalities algebraically.

10. Example 3: Solving Quadratic Equations by Using Algebra
Solve x2 – 10x + 18 ≤ –3 by using algebra.
Step 1 Write the related equation.
x2 – 10x + 18 = –3
Step 2 Solve for x to find the critical values.
x2 –10x + 21 = 0
(x – 3)(x – 7) = 0
x = 3 or x = 7
The critical values are 3 and 7. The critical values divide the number line into three intervals: x ≤ 3, 3 ≤ x ≤ 7, x ≥ 7.

11. Step 3 Test an x-value in each interval.
Example 3 Continued
Step 3 Test an x-value in each interval.
–3 –2 –
Critical values
Test points
x2 – 10x + 18 ≤ –3 x
(2)2 – 10(2) + 18 ≤ –3
Try x = 2.
(4)2 – 10(4) + 18 ≤ –3 
Try x = 4.
(8)2 – 10(8) + 18 ≤ –3
Try x = 8. x

12. Shade the solution regions on the number line.
Example 3 Continued
Shade the solution regions on the number line.
Use solid circles for the critical values because the inequality contains them.
The solution is 3 ≤ x ≤ 7 or [3, 7].
–3 –2 –

13. Example 4
Solve the inequality by using algebra.
x2 – 6x + 10 ≥ 2
x2 – 6x + 10 = 2
x2 – 6x + 8 = 0
(x – 2)(x – 4) = 0
x = 2 or x = 4
The critical values are 2 and 4.
The 3 intervals are:
x ≤ 2, 2 ≤ x ≤ 4, x ≥ 4.

14. Test an x-value in each interval.
Example 4
Test an x-value in each interval.
–3 –2 –
Critical values
Test points
x2 – 6x + 10 ≥ 2
(1)2 – 6(1) + 10 ≥ 2 
Try x = 1.
(3)2 – 6(3) + 10 ≥ 2 x
Try x = 3. 
(5)2 – 6(5) + 10 ≥ 2
Try x = 5.

15. Shade the solution regions on the number line.
Example 4
Use solid circles for the critical values because the inequality contains them.
Shade the solution regions on the number line.
The solution is x ≤ 2 or x ≥ 4.
–3 –2 –

16. Example 5: Problem-Solving Application
The monthly profit P of a small business that sells bicycle helmets can be modeled by the function P(x) = –8x x – 4200, where x is the average selling price of a helmet.
What range of selling prices will generate a monthly profit of at least $6000?

17. Example 5
Write the inequality.
–8x x – 4200 ≥ 6000
Find the critical values by solving the related equation.
–8x x – 4200 = 6000
Write as an equation.
–8x x – 10,200 = 0
Write in standard form.
Factor out –8 to simplify.
–8(x2 – 75x ) = 0

18. Example 5 or
x ≈ or x ≈ 48.96

19. Graph the critical points and test points.
Example 5
Graph the critical points and test points.
Critical values
Test points

20. Example 5
–8(25) (25) – 4200 ≥ 6000
Try x = 25.
5800 ≥ 6000 x
Try x = 45.
–8(45) (45) – 4200 ≥ 6000 
6600 ≥ 6000
–8(50) (50) – 4200 ≥ 6000
Try x = 50.
5800 ≥ 6000 x
The solution is approximately ≤ x ≤

21. the average price of a helmet needs to be
Example 5
For a profit of $6000,
the average price of a helmet needs to be
between $26.04 and $48.96, inclusive.

22. Lesson Quiz: Part I
1. Graph y ≤ x2 + 9x + 14.
Solve each inequality.
2. x2 + 12x + 39 ≥ 12
x ≤ –9 or x ≥ –3
3. x2 – 24 ≤ 5x
–3 ≤ x ≤ 8

23. Lesson Quiz: Part II
A boat operator wants to offer tours of San Francisco Bay. His profit P for a trip can be modeled by P(x) = –2x x – 788, where x is the cost per ticket.
What range of ticket prices will generate a profit of at least $500?
between $14 and $46, inclusive