1. Solving Inequalities Algebraically
Section P.6 – the last section of the chapter!!!

2. Guided Practice Solution: – 4 < x < 12 (– 4, 12)
Solve |x – 4| < 8 for x. Write your answer in interval
notation.
Solution:
– 4 < x < 12
(– 4, 12)

3. Solving Quadratic Inequalities
First, replace the inequality sign with an equal sign
Next, solve for the variable as with any other
quadratic equation
Finally, examine the graph to see which values
of x satisfy the original inequality (often a visual…)
You need to check the actual points for inclusion or not

4. Guided Practice Solution: Final Solution:2
Solve x – x – 12 > 0 for x. Write your answer in
interval notation.
Now graph:
Solution:
For what values of x is
the graph above zero?
Final Solution:

5. Guided Practice No Solution!!! Nothing below zero!2
Solve x + 2x + 2 < 0 for x. Write your answer in
interval notation.
Solve
Graph
For what values of x is this function below zero?
No Solution!!!
Nothing below zero!

6. Guided Practice:
Solve x + 2x – 1 > 0 graphically. 3 2
Solution:

7. Guided Practice:
Solve:
Solution:

8. How about a word problem?
Basic Projectile Motion
(if an object is propelled, and then only
subject to the force of gravity)
If an object is launched vertically from a point s feet above the
ground with an initial velocity of v ft/sec, then the vertical position
s (in feet) of the object t seconds after it is launched is:
s = –16t + v t + s 2

9. How about a word problem?
A rugby ball is kicked straight up from a height of 3 feet with
an initial velocity of 57 ft/sec.
When will the ball’s height above ground be 44 feet?
Solve:
At t = 1 sec (on the way up) and
at t = sec (on the way down)

10. How about another word problem?
A rugby ball is kicked straight up from a height of 3 feet with
an initial velocity of 57 ft/sec.
When will the ball’s height above ground be at least 44 feet?
At t between 1 and sec
[ 1 , ]

11. …and another word problem?
A projectile is launched straight up from ground level with an
initial velocity of 272 ft/sec.
1. When will the projectile’s height above ground be 960 ft?
Solve:
At t = 5 sec (on the way up) and
at t = 12 sec (on the way down)

12. Whiteboard Problem: Solution: [ – 4 , 5/2 ]
Solve 2x + 3x < 20 for x. Write your answer in
interval notation.
Solution:
[ – 4 , 5/2 ]

13. 8 8 Whiteboard Problem: Solution: ( – , 0.268] U [ 3.732 , )
Solve x – 4x + 1 > 0 for x. Write your answer in
interval notation.
Solution: 8
( – , 0.268] U [ , ) 8

14. Homework: p. 59 1-29 odd Whiteboard problem: Solution:
Solve :–3x > –2x – 6
Solution:
Homework: p odd