1.03E - Absolute Rule and Inverse

Absolute Value Functions ABSOLUTE VALUE FUNCTIONS 1.03E – FINDING RULE AND INVERSE It’s important to know the parts of the absolute and how they relate to parameters: • Parameter 𝑎: • Is the slope of the Right Branch • Is the opposite to the slope of the Left Branch • Parameter ℎ: • Is the midpoint of any two symmetrical points on the function • Parameter 𝑘: • Is the extreme of the function (max/min depending on sign of 𝑎) • Parameters (ℎ, 𝑘): • Are the vertex of the function • Are the POI of the two branches Key Information When finding the rule, always use STANDARD FORM: 𝑓 𝑥 =𝑎 𝑥−ℎ +𝑘 Key Information 1. 2. 3. 4. Substitute vertex in for (ℎ, 𝑘) Substitute point in for (𝑥, 𝑦) Solve for 𝑎 Summarize Given Vertex and a Point • Find the Rule of the Function: Vertex (0,2) Ex 1: Point (2, 3) Find the Rule of the Function: • ℎ, 𝑘 = 0,2 • 𝑥, 𝑦 = 2,3 𝑓 𝑥 =𝑎 𝑥−ℎ +𝑘 3=𝑎 2−0 +2 1=𝑎2 1=𝑎 2 1 =𝑎 2 Therefore: Ex 1: 𝑓 𝑥 = 1 2 𝑥 +2 • Find the Rule of the Function with vertex (3, 5) and passes through (5, 8): • 𝑓 𝑥 =𝑎 𝑥−ℎ +𝑘 • ℎ, 𝑘 = 3,5 • 𝑥, 𝑦 = (5,8) 8=𝑎 5−3 +5 3=𝑎2 3=𝑎 2 3 =𝑎 2 Therefore: Ex 2: 𝑓 𝑥 = 3 2 𝑥−3 +5 • Find the Rule of the Function with vertex (3, 5) and passes through (5, 8): • ℎ, 𝑘 = 3,5 • 𝑥, 𝑦 = (5,8) 3 ∆𝑦 8−5 = = 𝑠𝑙𝑜𝑝𝑒 = 2 5−3 ∆𝑥 𝑥 > ℎ, 𝑠𝑜 𝑎 = Therefore: NOTE 3 2 (if x > h, then a = slope) 𝑓 𝑥 = 3 2 𝑥−3 +5 • Find the Rule of the Function that passes through (1, 7), (5, 7) and (3, 1): Symmetrical • 𝑥 − 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝑣𝑒𝑟𝑡𝑒𝑥 1+5 ℎ= =3 2 ∴ 3,1 𝑖𝑠 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑒𝑥 • 𝑓 𝑥 =𝑎 𝑥−ℎ +𝑘 • ℎ, 𝑘 = 3,1 and 𝑥, 𝑦 = (1,7) 7=𝑎 1−3 +1 6 = 𝑎 −2 6=𝑎 2 3=𝑎 Therefore: 𝑓 𝑥 =3 𝑥−3 +1 Ex 3: • Find the Rule of the Function that passes through (1, 7), (5, 7) and (3, 1): • ℎ, 𝑘 = 3,1 and 𝑥, 𝑦 = (1,7) • ℎ, 𝑘 = 3,1 and 𝑥, 𝑦 = (5,7) ∆𝑦 1 − 7 6 𝑚= = = − = −3 ∆𝑥 3 − 1 2 ∆𝑦 1 − 7 −6 𝑚= = = =3 ∆𝑥 3 − 5 −2 Ex 3: NOTE • Find the Rule of the Function that passes through (6, −11) and (9, −17) and has a vertex with x-coordinate of 5 • ℎ=5 • 𝑓 𝑥 =𝑎 𝑥−5 +𝑘 −11 = 𝑎 6 − 5 + 𝑘 −17 = 𝑎 9 − 5 + 𝑘 −11 = 𝑎 + 𝑘 −17 = 4𝑎 + 𝑘 −11 − 𝑎 = −17 − 4𝑎 3𝑎 = −6 𝑎 = −2 substitute and solve for k 𝑘 = −9 Ex 5: 𝑓 𝑥 = −2 𝑥 − 5 − 9 • Find the Rule of the Function that passes through (6, −11) and (9, −17) and has a vertex with x-coordinate of 5 • ℎ=5 • Ask: are both points on same side? • 𝑎= −11− −17 6−9 = 6 −3 = −2 • 𝑓 𝑥 =𝑎 𝑥−ℎ +𝑘 −11 = −2 6 − 5 + 𝑘 −11 = −2 1 + 𝑘 −11 = −2 + 𝑘 −9 = 𝑘 𝑓 𝑥 = −2 𝑥 − 5 − 9 Ex 5 (alternate strategy) • Find the rule of the absolute value function that passes through the points (-10, -9), (1, -7), (-7, -3) and (-2, -1). • Question: which points are on which side of function? • Strategy: list points with increasing x-values, determine which points are on which side. • (-10, -9), (-7, -3), (-2, -1), (1, -7) • Now check slopes! Ex 6: • Find the rule of the absolute value function that passes through the points (-10, -9), (-7, -3), (-2, -1), (1, -7) 𝑎= 𝑎= −3−(−9) =2 −7−(−10) −1−(−3) 2 = −2−(−7) 5 (first two points) (middle two points) Not on same side! Check last two points: 𝑎= −7−(−1) 1−(−2) = −2 (first two pts on left side, last two on right) Find equation of each side, solve POI for vertex. Ex 6: • Find the rule of the absolute value function that passes through the points (-10, -9), (-7, -3), (-2, -1), (1, -7) • 𝑦 = 2𝑥 + 𝑏 • −9 = 2 −10 + 𝑏 • 𝑏 = 11 𝑦 = 2𝑥 + 11 (left side) • 𝑦 = −2𝑥 + 𝑏 • −7 = −2 1 + 𝑏 • 𝑏 = −5 Ex 6: 𝑦 = −2𝑥 − 5 (right side) • Find the rule of the absolute value function that passes through the points (-10, -9), (-7, -3), (-2, -1), (1, -7) • • • • • 2𝑥 + 11 = −2𝑥 − 5 4𝑥 = −16 𝑥 = −4 𝑦 = 2 −4 + 11 = 3 Function opens down POI = (-4, 3) = vertex a = -2 • 𝑓 𝑥 = −2 𝑥 + 4 + 3 Ex 6: • Jan recently opened her very own tea house! Over her first 10 months of business she recorded her profits. Opening day Jan made $1200 in profit. Business was slow for the first 3 months, steadily declining to a minimum value of 500 in profit. At the 3 month mark she hired a Public Relations and Marketing Guru and her profits began to climb steadily, at the opposite rate of the decline. • Find the function that represents this situation. Jan’s Tea • Two Points Given: 𝐴(0, 1200) 𝐵(3, 500) 500 − 1200 𝑎= 3−0 −700 𝑎= 3 700 𝑓 𝑥 = 𝑥 − 3 + 500 3 Jan’s Tea FINDING THE INVERSE Absolute Value Function • The inverse of the Basic Absolute Value Function is a RELATION 𝑓 𝑥 = 𝑥 𝑦≥0 𝑓 −1 𝑥 ⇒ 𝑥 = 𝑦 𝑦=𝑥 y= ± 𝑥, 𝑥 ≥ 0 Inverse 𝑦 = −𝑥 • 𝑓 𝑥 = −3 𝑥 + 1 + 4 ran 𝑓 = −∞, 4 • 𝑥 = −3 𝑦 + 1 + 4 𝑥 − 4 = −3 𝑦 + 1 𝑥−4 − = 𝑦+1 3 𝑥−4 𝑦+1=− 3 𝑦+1= 𝑥−4 𝑦=− −1 3 The Inverse (Pg 1 of 2) 𝑦= 𝑥−4 3 𝑥−4 −1 3 ran 𝑓 = −∞, 4 𝑥−4 𝑦=− −1 3 𝑥 4 𝑦 =− + −1 3 3 𝑥 1 𝑦1 = − + 3 3 dom 𝑓 −1 = −∞, 4 𝑥 𝑥−4 𝑦= −1 3 𝑥 4 𝑦 = − −1 3 3 𝑥 7 𝑦2 = − 3 3 1 −3+3 y= ቐ 𝑥 7 −3 3 𝑜𝑣𝑒𝑟 −∞, 4 The Inverse continued •Visions 1 •P53 3ac •P55: 12ab, 13, 18ef Homework