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TODAY IN ALGEBRA… Warm up: 5.2 Review-Writing an equation of a line given two points Learning Goal: 5.3 Writing a linear equation in Point-Slope form.

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Presentation on theme: "TODAY IN ALGEBRA… Warm up: 5.2 Review-Writing an equation of a line given two points Learning Goal: 5.3 Writing a linear equation in Point-Slope form."β€” Presentation transcript:

1 TODAY IN ALGEBRA… Warm up: 5.2 Review-Writing an equation of a line given two points Learning Goal: 5.3 Writing a linear equation in Point-Slope form Independent Practice

2 WARM UP: Write an equation for the linear function 𝑓 with the values 𝑓 βˆ’2 =10 and 𝑓 4 =βˆ’2
Rewrite function values as coordinate pairs (points): 2. Calculate slope: π‘š= π‘š= βˆ’12 6 Calculate the y-intercept: 𝑦=π‘šπ‘₯+𝑏 βˆ’2=βˆ’2 4 +𝑏 βˆ’2=βˆ’8+𝑏 πŸ”=𝒃 βˆ’πŸ, 𝟏𝟎 , (πŸ’, βˆ’πŸ) π‘₯ 𝑦 βˆ’2βˆ’10 REMEMBER: 𝒇(𝒙)=π’Žπ’™+𝒃 4βˆ’(βˆ’2) =βˆ’πŸ π‘šβ†’ π‘ π‘™π‘œπ‘π‘’: π‘¦βˆ’π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘: βˆ’2 6 𝑏→ 𝑓(π‘₯)= π‘₯+ βˆ’2 6

3 New…POINT-SLOPE FORM:
5.3 WRITING LINEAR EQUATIONS IN POINT-SLOPE FORM We learned… SLOPE-INTERCEPT FORM: New…POINT-SLOPE FORM: π‘¦βˆ’ 𝑦 1 =π‘š(π‘₯βˆ’ π‘₯ 1 ) 𝑦=π‘šπ‘₯+𝑏 π‘ π‘™π‘œπ‘π‘’= π‘Ÿπ‘–π‘ π‘’ π‘Ÿπ‘’π‘› π‘π‘œπ‘–π‘›π‘‘: ( π‘₯ 1 , 𝑦 1 ) π‘ π‘™π‘œπ‘π‘’= π‘Ÿπ‘–π‘ π‘’ π‘Ÿπ‘’π‘› π‘¦βˆ’π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘: (0, 𝑏) π‘ π‘™π‘œπ‘π‘’= 1 3 𝑔𝑖𝑣𝑒𝑛 π‘π‘œπ‘–π‘›π‘‘:(4, 2) π‘¦βˆ’ = (π‘₯βˆ’ ) 1 3 2 4

4 π‘¦βˆ’ = π‘₯βˆ’ 𝑦+3=2(π‘₯βˆ’4) π‘¦βˆ’ = π‘₯βˆ’ π‘¦βˆ’4=βˆ’2(π‘₯+1) βˆ’3 2 4 4 βˆ’2 βˆ’1
5.3 WRITING LINEAR EQUATIONS IN POINT-SLOPE FORM PRACTICE: Write an equation of the line in point-slope form given a point and slope. REMEMBER: π’šβˆ’ π’š 𝟏 =π’Ž(π’™βˆ’ 𝒙 𝟏 ) 1. π‘ π‘™π‘œπ‘π‘’: π‘ π‘™π‘œπ‘π‘’:βˆ’2 π‘π‘œπ‘–π‘›π‘‘:(4, βˆ’3) π‘π‘œπ‘–π‘›π‘‘:(βˆ’1, 4) π‘¦βˆ’ = π‘₯βˆ’ 𝑦+3=2(π‘₯βˆ’4) βˆ’3 2 4 π‘¦βˆ’ = π‘₯βˆ’ π‘¦βˆ’4=βˆ’2(π‘₯+1) 4 βˆ’2 βˆ’1

5 On a separate piece of paper try the following problems:
PRACTICE: (10 MINUTES) On a separate piece of paper try the following problems: Pg. 305: 3-11

6 1. 𝑦+3=2 π‘₯βˆ’4 𝑦+3=2π‘₯βˆ’8 βˆ’ 3 βˆ’ 3 π’š=πŸπ’™βˆ’πŸπŸ π‘ π‘™π‘œπ‘π‘’: π‘¦βˆ’π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘:
5.3Changing POINT-SLOPE FORM to SLOPE-INTERCEPT FORM Rewrite the equation into slope-intercept form. SOLVE FOR Y! 1. 𝑦+3=2 π‘₯βˆ’4 𝑦+3=2π‘₯βˆ’8 βˆ’ βˆ’ 3 π’š=πŸπ’™βˆ’πŸπŸ π‘ π‘™π‘œπ‘π‘’: π‘¦βˆ’π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘: 2. π‘¦βˆ’4=βˆ’2 π‘₯+1 π‘¦βˆ’4=βˆ’2π‘₯βˆ’2 π’š=βˆ’πŸπ’™+𝟐 π‘ π‘™π‘œπ‘π‘’: π‘¦βˆ’π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘: 2 βˆ’2 βˆ’11 2

7 𝑦+2= 2 3 (π‘₯βˆ’3) 2 3 π‘ π‘™π‘œπ‘π‘’: π‘π‘œπ‘–π‘›π‘‘: (3, βˆ’2)
5.3 Graphing equations in POINT-SLOPE FORM **Points are always opposite the sign!!! Graph the equation: π‘π‘œπ‘–π‘›π‘‘:( π‘₯ 1 , 𝑦 1 ) REMEMBER: π’šβˆ’ π’š 𝟏 =π’Ž(π’™βˆ’ 𝒙 𝟏 ) 𝑦+2= 2 3 (π‘₯βˆ’3) π‘ π‘™π‘œπ‘π‘’= π‘Ÿπ‘–π‘ π‘’ π‘Ÿπ‘’π‘› 3 2 π‘‘π‘œ 𝑔𝑒𝑑 π‘‘π‘œ 𝑛𝑒π‘₯𝑑 π‘π‘œπ‘–π‘›π‘‘β†’ 2 3 π‘ π‘™π‘œπ‘π‘’: π‘π‘œπ‘–π‘›π‘‘: (3, βˆ’2) π‘ π‘‘π‘Žπ‘Ÿπ‘‘β†’

8 π‘¦βˆ’2= 1 2 (π‘₯+2) 1 2 π‘ π‘™π‘œπ‘π‘’: π‘π‘œπ‘–π‘›π‘‘: (βˆ’2, 2)
5.3 Graphing equations in POINT-SLOPE FORM **Points are always opposite the sign!!! PRACTICE: Graph the equation: π‘π‘œπ‘–π‘›π‘‘:( π‘₯ 1 , 𝑦 1 ) REMEMBER: π’šβˆ’ π’š 𝟏 =π’Ž(π’™βˆ’ 𝒙 𝟏 ) π‘¦βˆ’2= 1 2 (π‘₯+2) π‘ π‘™π‘œπ‘π‘’= π‘Ÿπ‘–π‘ π‘’ π‘Ÿπ‘’π‘› 2 π‘‘π‘œ 𝑔𝑒𝑑 π‘‘π‘œ 𝑛𝑒π‘₯𝑑 π‘π‘œπ‘–π‘›π‘‘β†’ 1 2 1 π‘ π‘™π‘œπ‘π‘’: π‘π‘œπ‘–π‘›π‘‘: (βˆ’2, 2) π‘ π‘‘π‘Žπ‘Ÿπ‘‘β†’

9 On the same piece of paper try the following problems:
PRACTICE: (15 MINUTES) On the same piece of paper try the following problems: Pg. 306: 14-19

10 5.3 Writing equations in POINT-SLOPE FORM given a graph
Write an equation in point-slope form of the line shown. REMEMBER: π’šβˆ’ π’š 𝟏 =π’Ž(π’™βˆ’ 𝒙 𝟏 ) 1. Find slope: π‘ π‘™π‘œπ‘π‘’= π‘Ÿπ‘–π‘ π‘’ π‘Ÿπ‘’π‘› π‘š=βˆ’1 Choose one point: *doesn’t matter which one* (1, 1) 3. Plug into point-slope form. (βˆ’1, 3) = βˆ’2 2 βˆ’2 (1, 1) 2 π‘¦βˆ’ = π‘₯βˆ’ π‘¦βˆ’1=βˆ’(π‘₯βˆ’1) 1 βˆ’1 1

11 5.3 Writing equations in POINT-SLOPE FORM given points
Write an equation in point-slope form given the two points. REMEMBER: π’šβˆ’ π’š 𝟏 =π’Ž(π’™βˆ’ 𝒙 𝟏 ) 𝟐, πŸ‘ , (πŸ’, πŸ’) 1. Find slope: π‘š= π‘š= 1 2 Choose one point. *doesn’t matter which one* 3. Plug into point-slope form. 4βˆ’3 1 2 4βˆ’2 π‘¦βˆ’ = π‘₯βˆ’ 𝑂𝑅 3 2 (𝟐, πŸ‘) οƒ  1 2 4 4 (πŸ’, πŸ’)οƒ 

12 HOMEWORK #4: Pg. 305: 3-30 all If finished, work on other assignments:
HW #1: Pg. 286: 3-23 all HW #2: Pg. 286: all HW #3: Pg. 296: 3-7, 11-14, 22-25


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