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Chapter P: Prerequisite InformationPowerPoint Presentation

Chapter P: Prerequisite Information

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Presentation Transcript

Objectives

- You will learn about:
- Solving absolute value inequalities
- Solving quadratic inequalities
- Approximating solutions to inequalities
- Projectile motion

- Why:
- These techniques are involved in using a graphing utility to solve inequalities

Vocabulary

- Union of two sets A and B
- Projectile motion

Solving Absolute Value Inequalities

- Let u be an algebraic expression in x and let a be a real number with a ≥ 0.
- If |u| < a, then u is in the interval (-a, a).
- That is, |u| < a if and only if –a < u < a

- If |u| > a, then u is in the interval (-∞, -a) or (a, ∞).
- That is |u| > a if and only if u < -a or u > a

- If |u| < a, then u is in the interval (-a, a).

Example 1:Solve an Absolute Value Inequality

- Solve |x – 4|< 8

Example 2:Solving another absolute value inequality

- Solve |3x – 2|≥ 5

Example 3:Solving a Quadratic Inequality

- Solve x2 – x – 12 > 0
- First, solve the equation x2 – x – 12 = 0
- Graph the equation and observe where the graph is above zero.

Example 4:Solving another quadratic inequality

- Solve 2x2 + 3x ≤ 20
- Again, solve the equation and graph

Example 5:Solving another quadratic inequality

- Solve x2– 4x + 1 > 0
- This one we must do graphically because the equation does not factor.
- Enter on your calculator: y = x2 – 4x + 1
- We will use the trace key to observe where we have zeros.

Example 6:Showing there is no solution

- Solve x2+ 2x + 2 ≤ 0.
- Graph the equation. Where is the equation below the x-axis?
- Use the quadratic formula to verify your answer.

Example 7:Solving a cubic inequality

- Solve x3 +2x2 + 2 ≥ 0
- Enter the equation in y =
- Estimate where your zeros are and then use the zero trace function to find the values.

Projectile Motion

- Suppose an object is launched vertically from a point s0 feet above the ground with an initial velocity v0 feet per second. The vertical position s (in feet) of the object t seconds after it is launched is:
- s = -16t2 + v0t + s0

Example 8:Finding the height of a projectile

- A projectile is launched straight up from ground level with an initial velocity of 288 ft/sec.
- When will the projectile’s height above the ground be1152 ft?
- When will the projectile’s height above the ground be at least 1152 ft?

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