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Chapter P: Prerequisite Information. Section P-7: Solving Inequalities Algebraically and Graphically. Objectives. You will learn about: Solving absolute value inequalities Solving quadratic inequalities Approximating solutions to inequalities Projectile motion Why:
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Chapter P:Prerequisite Information Section P-7: Solving Inequalities Algebraically and Graphically
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
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?