1.03C - Absolute Equations | Lecture Note

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High School
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High School Physics
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2024
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Absolute Value Functions Unit 1 - Functions ABSOLUTE VALUE FUNCTIONS 1.03C – SOLVING EQUATIONS • Follow BEDMAS with absolute value falling under B • To drop absolute brackets, you must make positive: • Ex 1: 4 3 𝑓 𝑥 = −2 − 𝑥 + 9 4 −7 • Find 𝑓 12 4 3 𝑓 12 = −2 − 12 + 9 4 =− 55 3 Finding 𝑦 given 𝑥 −7 • Ex 3: 1 𝑓 𝑥 = − 2𝑥 + 3 − 7 4 • Find initial value 1 𝑓 0 =− 2 0 +3 −7 4 31 =− 4 Finding initial value 𝑥 =5 𝑥 = −5 𝑥=5 ∴ 𝑥 = {−5, 5} The Basics of Solving For 𝑥 1. Set Equal to Zero 2. Isolate the Concept 𝑏 𝑥 − ℎ 3. Make two cases • Negative Case • Positive Case 𝑏 𝑥 − ℎ = −_______ 𝑏 𝑥 − ℎ =_______ 4. Solve for 𝑥 5. Validate both! 6. Summarize Finding Zeros - Steps 𝑓 𝑥 = 𝑥 −6 𝑥 −6=0 Check 𝑥 =6 Check 6 −6=0 −6 − 6 = 0 6 −6=0 𝑥 = −6 𝑥=6 6 −6= 0 0=0 0=0 ∴ 𝑥 = −6, 6 Find the Zeros 𝑓 𝑥 =2 𝑥+3 −2 Check Check 2 (−2) + 3 − 2 = 0 2 𝑥+3 −2=0 𝑥+3 =1 2 1 −2=0 2(1) − 2 = 0 2 (−4) + 3 − 2 = 0 2 −1 − 2 = 0 2(1) − 2 = 0 𝑥 + 3 = −1 𝑥+3=1 2−2=0 𝑥 = −1 − 3 𝑥 =1−3 2−2=0 0=0 𝑥 = −4 𝑥 = −2 0=0 ∴ 𝑥 = −4, −2 Find the Zeros – By Algebra 𝑓 𝑥 =2 𝑥−5 +6 2 𝑥−5 +6=0 𝑥−5 =− 6 2 𝑁𝑜𝑡 𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒‼ 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑚𝑎𝑘𝑒𝑠 𝑡ℎ𝑖𝑛𝑔𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 ∴ No Zeros Find the Zeros 1. Set Equal to Zero 2. Isolate the Concept 𝑏 𝑥 − ℎ 3. Make two cases • Negative Case • Positive Case 𝑏 𝑥 − ℎ = −_______ 𝑏 𝑥 − ℎ =_______ 4. Solve for 𝑥 5. Test 6. Summarize Solving Equations- Steps − 𝑥 − 2 + 4 = −2 − 𝑥 − 2 = −6 𝑥−2 =6 Check − −4 − 2 + 4 = −2 − −6 + 4 = −2 − 6 + 4 = −2 −2 = −2 Check − 8 − 2 + 4 = −2 𝑥 − 2 = −6 𝑥 = −4 𝑥−2=6 𝑥=8 ∴ 𝑥 = −4, 8 Example 1 − 6 + 4 = −2 − 6 + 4 = −2 −2 = −2 − 𝑥 − 2 + 4 = −2 Vertex: 2, 4 Parameter 𝑎 = −1 1 𝑚= 1 Since it is negative, then the vertex is a maximum Line 2: y = −2 By Graphing 2 𝑥−5 +6=2 2 𝑥 − 5 = −4 𝑥 − 5 = −2 𝑥= ∅ Example 2 2 𝑥−5 +6=2 Vertex: 5,6 Parameter 𝑎 = 2 2 𝑚= 1 Since it is positive, then the vertex is a minimum Line 2: y = 2 Check the Graph 𝑥 − 2 = 1 − 2𝑥 Check 𝑥 − 2 = −(1 − 2𝑥) −𝑥 = 1 𝑥 = −1 −1 − 2 = 1 − 2(−1) 𝑥 − 2 = 1 − 2𝑥 3𝑥 = 3Check 𝑥=1 𝑥 = −1, 1 −3 = 1 + 2 1 − 2 = 1 − 2(1) −1 = 1 − 2 1 = −1 3 =3 𝑥 = −1 Example 3 𝐸𝑥𝑡𝑟𝑎𝑛𝑒𝑜𝑢𝑠 𝑥 − 2 = 1 − 2𝑥 Vertex: 2,0 Parameter 𝑎 = 1 1 𝑚= 1 Since it is positive, then the vertex is a minimum Line 2: y = −2𝑥 + 1 Check the Graph -2 𝑥 + 1 + 9 = 3𝑥 + 2 Try this! 2 𝑥 − 2 + 6 = 2𝑥 + 2 2 𝑥 − 2 = 2𝑥 − 4 𝑥−2 =𝑥−2 𝑥−2=− 𝑥−2 2𝑥 = 4 𝑥=2 Example 4 𝑥−2=𝑥−2 0𝑥 = 0 ?? 2 𝑥 − 2 + 6 = 2𝑥 + 2 Vertex: 2,6 Parameter 𝑎 = 2 2 𝑚= 1 Since it is positive, then the vertex is a minimum Line 2: y = 2𝑥 + 2 Graph 2 𝑥 − 2 + 6 = 2𝑥 + 2 2 𝑥 − 2 = 2𝑥 − 4 𝑥−2 =𝑥−2 𝑥−2=− 𝑥−2 𝑥−2=𝑥−2 2𝑥 = 4 𝑥=2 0𝑥 = 0 ∴ 𝑥 = [2, ∞[ Example 4 𝑥−2= 𝑥−2 +4 Check 4−2= 4−2 +4 𝑥−6= 𝑥−2 𝑥−6= 𝑥−2 2= 2 +4 2= 2 +4 2≠6 𝑥−2=𝑥−6 0 ≠ −4 𝑬𝒙𝒕𝒓𝒂𝒏𝒆𝒐𝒖𝒔 Solve 𝑥 − 2 = −(𝑥Extraneous − 6) 𝑥 − 2 = −𝑥 + 6 2𝑥 = 8 𝑥=4 ∴𝑥= ∅ 𝑥−2= 𝑥−2 +4 Vertex: 2,4 Parameter 𝑎 = 1 1 𝑚= 1 Since it is positive, then the vertex is a minimum Line 2: y = 𝑥 − 2 Graph 5 𝑥+1 =2 𝑥+1 +6 Challenge 1 2𝑥 + 1 = 5 𝑥 − 3 Challenge 2 •P54: #6a,c,e,g,i •P56: #17a,b (challenge!) Homework