1.02B - Square Root Transformed and Graphing

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2024
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SQUARE ROOT FUNCTIONS SQUARE ROOT FUNCTIONS 1.02B – TRANSFORMED FUNCTION USING ALL PARAMETERS: 𝑓 𝑥 =𝑎 𝑏 𝑥−ℎ +𝑘 • The transformed square root function can be simplified • However, you cannot square root a negative, and the parameter 𝑏 can have a positive or a negative value: 𝑓 𝑥 = 𝑎 ± 𝑏 (𝑥 − ℎ) + 𝑘 because 𝑎𝑏 = 𝑎 ⋅ 𝑏 𝑓 𝑥 = 𝑎 𝑏 ±1(𝑥 − ℎ) + 𝑘 Considering 𝑎 𝑏 the new value of parameter 𝑎, we get: 𝑓 𝑥 = 𝑎𝑛𝑒𝑤 ±1(𝑥 − ℎ) + 𝑘 Simplifying Parameter 𝑏 𝑓 𝑥 = 𝑎 (𝑥 − ℎ) + 𝑘 OR 𝑓 𝑥 = 𝑎 −(𝑥 − ℎ) + 𝑘 The Transformed Rule 𝑓 𝑥 = 𝑎 (𝑥 − ℎ) + 𝑘 OR 𝑓 𝑥 = 𝑎 −(𝑥 − ℎ) + 𝑘 The Transformed Rule • This function always has restrictions • −4 fails because −4 < 0 radicand Restriction: 𝑟𝑎𝑑𝑖𝑐𝑎𝑛𝑑 ≥ 0 𝑏 𝑥−ℎ ≥0 New Step 𝑓 𝑥 = 3 −4𝑥 + 8 − 3 = 3 −4 (𝑥 − 2) − 3 = 3 4 −(𝑥 − 2) − 3 = 3(2) −(𝑥 − 2) − 3 = 6 −(𝑥 − 2) − 3 Example 1: simplify b 𝑓 𝑥 = 3 2𝑥 − 8 − 1 = 3 2(𝑥 − 4) − 1 = 3 2 (𝑥 − 4) − 1 Example 2: Simplify b 𝑎>0 𝑏 = −1 𝑎>0 𝑏=1 𝑎<0 𝑎<0 𝑏 = −1 𝑏 = 1 The options … • The curve is in the shape of a • • • • swish: swish upwards and right from vertex if 𝑎 > 0 and 𝑏 = 1 swish downwards and right from vertex if 𝑎 < 0 and 𝑏 = 1 swish upwards and left from vertex if 𝑎 > 0 and 𝑏 = −1 swish downwards and left from vertex if 𝑎 < 0 and 𝑏 = −1 • The vertex is (ℎ, 𝑘) Remember PROPERTIES If 𝑓 𝑥 = 𝑎 𝑥 − ℎ + 𝑘 (ℎ, 𝑘) 𝑎<0 𝑎>0 𝑏=1 𝑎<0 𝑎>0 𝑑𝑜𝑚𝑓 = [ℎ, +∞[ 𝑎<0 𝑎>0 𝑟𝑎𝑛 =] − ∞, 𝑘] 𝑟𝑎𝑛 = [𝑘, +∞[ 𝑎<0 Initial Value ⇒ 𝑓(0) 𝑎>0 There will only ever be ONE zero There will only be a zero if 1) 𝑘 = 0 2) 𝑎 and 𝑘 have opposite signs Zeros 𝑎<0 𝑎>0 𝑆𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑆𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 Variation 𝑎<0 max 𝑓 = 𝑘 Extremes 𝑎>0 min 𝑓 = 𝑘 Changes at the zero -ORIf there is no zero … Will always be the same Sign 𝑎<0 𝑎>0 𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 Sign - If No Zeros … Domain Range Intercepts Variation Extremes Sign 𝑑𝑜𝑚 𝑓 = [𝒉, ∞[ 𝑟𝑎𝑛 𝑓 = [𝑘, +∞[ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 ⇒ 𝑓(0) 𝑧𝑒𝑟𝑜𝑠 There will only be a zero if 1) 𝑘 = 0 2) 𝑎 and 𝑘 have opposite signs 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 [ℎ, +∞[ min 𝑓 = 𝑘 max 𝑓 = 𝐷𝑁𝐸 If no zeros: strictly positive over[ℎ, +∞[ If a zero exists at 𝑥1 : negative over ℎ, 𝑥1 positive over [𝑥1 , ∞[ Domain Range Intercepts Variation Extremes Sign 𝑑𝑜𝑚 𝑓 = [𝒉, ∞[ 𝑟𝑎𝑛 𝑓 =] − ∞, 𝑘] 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 ⇒ 𝑓(0) 𝑧𝑒𝑟𝑜𝑠 There will only be a zero if 1) 𝑘 = 0 2) 𝑎 and 𝑘 have opposite signs 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 [ℎ, +∞[ min 𝑓 = 𝐷𝑁𝐸 max 𝑓 = 𝑘 If no zeros: strictly negative over[ℎ, +∞[ If a zero exists at 𝑥1 : positive over ℎ, 𝑥1 negative over [𝑥1 , ∞[ PROPERTIES If 𝑓 𝑥 = 𝑎 −(𝑥 − ℎ) + 𝑘 (ℎ, 𝑘) 𝑎<0 𝑎>0 𝑏 = −1 𝑎<0 𝑎>0 𝑑𝑜𝑚 𝑓 =] − ∞, ℎ] 𝑎<0 𝑟𝑎𝑛 =] − ∞, 𝑘] 𝑎>0 𝑟𝑎𝑛 = [𝑘, +∞[ 𝑎<0 Initial Value ⇒ 𝑓(0) 𝑎>0 There will only ever be ONE zero There will only be a zero if 1) 𝑘 = 0 2) 𝑎 and 𝑘 have opposite signs Zeros 𝑎<0 𝑆𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 Variation 𝑎>0 𝑆𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑎<0 max 𝑓 = 𝑘 Extremes 𝑎>0 min 𝑓 = 𝑘 Changes at the zero -ORIf there is no zero … Will always be the same Sign 𝑎<0 𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒 Sign – If No Zeros … 𝑎>0 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 Domain Range Intercepts Variation Extremes Sign 𝑑𝑜𝑚 𝑓 =] − ∞, 𝒉] 𝑟𝑎𝑛 𝑓 = [𝑘, +∞[ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 ⇒ 𝑓(0) 𝑧𝑒𝑟𝑜𝑠 There will only be a zero if 1) 𝑘 = 0 2) 𝑎 and 𝑘 have opposite signs 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 ] − ∞, ℎ] min 𝑓 = 𝑘 max 𝑓 𝐷𝑁𝐸 If no zeros: strictly positive over ] − ∞, ℎ] If a zero exists at 𝑥1 : positive over ] − ∞, 𝑥1 ] negative over [𝑥1 , ℎ] Domain Range Intercepts Variation Extremes Sign 𝑑𝑜𝑚 𝑓 =] − ∞, 𝒉] 𝑟𝑎𝑛 𝑓 =] − ∞, 𝑘] 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 ⇒ 𝑓(0) 𝑧𝑒𝑟𝑜𝑠 There will only be a zero if 1) 𝑘 = 0 2) 𝑎 and 𝑘 have opposite signs 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 ] − ∞, ℎ] min 𝑓 = 𝐷𝑁𝐸 max 𝑓 = 𝑘 If no zeros: strictly negative over] − ∞, ℎ] If a zero exists at 𝑥1 : negative over ] − ∞, 𝑥1 ] positive over [𝑥1 , ℎ] 𝑓 𝑥 = 3 −𝑥 − 3 − 6 = 3 −(𝑥 + 3) − 6 , 𝑥 ≤ −3 Find the Properties 𝑓 𝑥 = 3 −(𝑥 + 3) − 6, 𝑥 ≤ −3 𝑎 = 3, 𝑏 = −1, ℎ = −3, 𝑘 = −6 Find the Properties Domain Range Intercepts Variation Extremes Sign 𝑑𝑜𝑚 𝑓 =] − ∞, 𝒉] 𝑟𝑎𝑛 𝑓 = [𝑘, +∞[ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 ⇒ 𝑓(0) 𝑧𝑒𝑟𝑜𝑠 There will only be a zero if 1) 𝑘 = 0 2) 𝑎 and 𝑘 have opposite signs 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 ] − ∞, ℎ] min 𝑓 = 𝑘 max 𝑓 𝐷𝑁𝐸 If no zeros: strictly positive over ] − ∞, ℎ] If a zero exists at 𝑥1 : positive over ] − ∞, 𝑥1 ] negative over [𝑥1 , ℎ] Domain 𝑑𝑜𝑚 𝑓 =] − ∞, −𝟑] Range 𝑟𝑎𝑛 𝑓 = [−6, +∞[ Intercepts 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 ⇒ 𝐷𝑁𝐸 𝑧𝑒𝑟𝑜𝑠 ⇒ 𝑥 = {−7} Variation 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑜𝑣𝑒𝑟 ] − ∞, −3] Extremes min 𝑓 = −6 max 𝑓 𝐷𝑁𝐸 Sign positive over ] − ∞, −7] negative over [−7, −3] 1. Ensure that parameter 𝑏 is factored away from 𝑥−ℎ 2. Apply the parameters to the Ordered Pair Formula 𝒙 + 𝒉, 𝒂𝒚 + 𝒌 𝒃 3. Transform the 4 basic pairs, one at a time Graphing Square Roots • Basic Ordered Pairs: (𝑥, 𝑦) 0,0 1,1 4, 2 (9, 3) To Graph • 𝑓 𝑥 = 3 4𝑥 − 8 − 1 = 3 4(𝑥 − 2) − 1 𝑥 + ℎ, 𝑎𝑦 + 𝑘 𝑏 0,0 ↦ 1,1 ↦ 4, 2 ↦ (9, 3) ↦ Graph 𝑥 + 2, 3𝑦 − 1 4 2, −1 9 ,2 4 3, 5 17 ,8 4 • 𝑓 𝑥 = 3 4(𝑥 − 2) − 1 • 0,0 ↦ 2, −1 • 1,1 ↦ 9 ,2 4 • 4,2 ↦ (3,5) • 9,3 ↦ 17 ,8 4 Graph 𝑓 𝑥 = 3 4𝑥 − 8 − 1 • 𝑓 𝑥 = 3 4𝑥 − 8 − 1 = 3 4(𝑥 − 2) − 1 = 3 4 (𝑥 − 2) − 1 = 3(2) (𝑥 − 2) − 1 = 6 (𝑥 − 2) − 1 Graph but simplify first • 𝑓 𝑥 = 6 (𝑥 − 2) − 1 • 𝑥/𝑏 + ℎ, 𝑎𝑦 + 𝑘 ↦ (𝑥 + 2, 6𝑦 − 1) • • • • 0,0 1,1 4,2 9,3 ↦ 0 + 2, ↦ 1 + 2, ↦ (4 + 2, ↦ (9 + 2, 6 6 6 6 0 1 2 3 − 1 ↦ 2, −1 − 1 ↦ 3,5 − 1) ↦ (6,11) − 1) ↦ (11, 17) • 𝑓 𝑥 = 6 (𝑥 − 2) − 1 • • • • 0,0 1,1 4,2 9,3 ↦ 2, −1 ↦ 3,5 ↦ (6,11) ↦ (11, 17) 𝑔 𝑥 = 2 −9𝑥 + 18 − 3 Try It: Find Ordered Pairs Graph 𝑓 𝑥 = 2 − 9 − 𝑥 𝑓 𝑥 = − −𝑥 + 9 + 2 𝑓 𝑥 = − −(𝑥 − 9) + 2 𝑎 = −1, 𝑏 = −1, ℎ = 9, 𝑘 = 2 Try this! 𝑓 𝑥 = − −(𝑥 − 9) + 2 • −𝑥 + ℎ, 𝑎𝑦 + 𝑘 ↦ (−𝑥 + 9, −1 𝑦 + 2) • • • • 0,0 1,1 4,2 9,3 ↦ (−0 + 9, ↦ (−1 + 9, ↦ (−4 + 9, ↦ (−9 + 9, Try this! −1 (0) + 2) ↦ 9, 2 −1 (1) + 2) ↦ 8, 1 −1 (2) + 2) ↦ (5,0) −1 (3) + 2) ↦ (0, −1) 𝑓 𝑥 = − −(𝑥 − 9) + 2 • • • • Vertex 0,0 1,1 4,2 9,3 ↦ 9, 2 ↦ 8, 1 ↦ (5,0) ↦ (0, −1) Try this! •Page 40-41 •#6, 7, 10 Homework: Visions 1