Aim: How can we review for test # 3? Test Part I Tomorrow Begin Now: Complete worksheet on domain. Also determine the range for each function.

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Aim: How can we review for test # 3? Test Part I Tomorrow Begin Now: Complete worksheet on domain. Also determine the range for each function

Aim: How can we review for test # 3? Test Part I Tomorrow 1. Evaluate: 2. Simplify: 3. Simplify the difference quotient: 4. Determine the domain of I. Using the function II. Solve the following inequalities: 1. 2a. 3. 2b.

Aim: How can we review for test # 3? Test Part I Tomorrow III. 1. An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = – 4.9t ² t , where s is in meters. When does the object strike the ground? 2. An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object's maximum height? When will it attain this height?

Aim: How can we review for test # 3? Test Part I Tomorrow Solutions I. 1.

Aim: How can we review for test # 3? Test Part I Tomorrow Solutions III 1. 0 = – 4.9t ² t = t – 4t – 12 0 = (t – 6)(t + 2) Then t = 6 or t = 2. t = –2 is an extraneous solution The object strikes the ground six seconds after launch. s(t) = –16t ² + 64t + 80 h = –b/2a = –(64)/2(–16) = –64/–32 = 2 k = s(2) = –16(2)2+ 64(2) + 80 = –16(4) = 208 – 64 = 144 the vertex is at (2, 144) It takes two seconds to reach the maximum height of 144 feet. 2.