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U1B L2 Reviewing Linear Functions

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1 U1B L2 Reviewing Linear Functions
UNIT 1B LESSON 2 REVIEW OF LINEAR FUNCTIONS

2 U1B L2 Reviewing Linear Functions
Equations of Lines The vertical line through the point (a, b) has equation x = a since every x-coordinate on the line has the same value a. 𝟐,πŸ“ βˆ’πŸ,πŸ‘ 𝟎,πŸ‘ 𝟐,πŸ‘ πŸ’,πŸ‘ 𝟐,πŸ‘ Similarly, the horizontal line through (a, b) has equation y = b 𝟐,𝟎 The horizontal line through the point (2, 3) has equation 𝟐,βˆ’πŸ y = 3 The vertical line through the point (2, 3) has equation x = 2

3 Finding Equations of Vertical and Horizontal Lines
U1B L2 Reviewing Linear Functions Finding Equations of Vertical and Horizontal Lines EXAMPLE 1 Write the equations of the vertical and horizontal lines through the point βˆ’πŸ‘, πŸ– Horizontal Line is y = 8 βˆ’πŸ‘, πŸ– Vertical Line is x = – 3

4 U1B L2 Reviewing Linear Functions
EXAMPLE 2: Reviewing Slope-Intercept Form of Linear Functions Y1 = 2x + 7 Slope y-intercept form y = mx + b slope y-intercept (0, b) x Y = 2x + 7 βˆ’3 𝟐 βˆ’πŸ‘ +πŸ•=𝟏 𝟐 𝟎 +πŸ•=πŸ• y – intercept ( , ) 𝟎 πŸ• (𝟎,πŸ•) π’Ž= π’“π’Šπ’”π’† 𝒓𝒖𝒏 = π’š 𝟐 βˆ’ π’š 𝟏 𝒙 𝟐 βˆ’ 𝒙 𝟏 π’Ž= πŸβˆ’πŸ• βˆ’πŸ‘βˆ’πŸŽ = βˆ’πŸ” βˆ’πŸ‘ =𝟐 (βˆ’πŸ‘,𝟏)

5 U1B L2 Reviewing Linear Functions
Unit 1B Lesson 2 Page 1 EXAMPLES State the slopes and y-intercepts of the given linear functions. y = 4x slope = m = _______ y -intercept ( , ) 3. 4 0 , 0 y = 3x – slope = m = _______ y -intercept ( , ) 4. 3 0 , βˆ’πŸ“ = slope = m = _______ y -intercept ( , ) 6. β…“ 0 , βˆ’πŸ slope = m = _______ y -intercept ( , ) 6. βˆ’ 𝟏 𝟐 0 , Β½ 𝒇 𝒙 = 𝟏 𝟐 βˆ’ 𝟏 𝟐 𝒙

6 General Linear Equation
U1B L2 Reviewing Linear Functions General Linear Equation Although the general linear form helps in the quick identification of lines, the slope-intercept form is the one to enter into a calculator for graphing. Ax + By = C By = – Ax + C y = – (A/B) x + C/B

7 Analyzing and Graphing a General Linear Equation
U1B L2 Reviewing Linear Functions Analyzing and Graphing a General Linear Equation Example 7 Find the slope and y-intercept of the line 2π‘₯βˆ’3𝑦=15 Rearrange for y βˆ’πŸ‘π’š = βˆ’πŸπ’™ + πŸπŸ“ βˆ’πŸ‘ βˆ’πŸ‘ π’š = βˆ’πŸ βˆ’πŸ‘ 𝒙+ πŸπŸ“ βˆ’πŸ‘ πŸ” π’š= 𝟐 πŸ‘ π’™βˆ’πŸ“ πŸ‘ πŸ’ 𝟐 y-intercept is (𝟎, βˆ’πŸ“) Slope is 𝟐 πŸ‘

8 U1B L2 Reviewing Linear Functions
Unit 1B Lesson 2 Page 1 EXAMPLES State the slopes and y-intercepts of the given linear functions. βˆ’ 𝟏 𝟐 x + 2y = 3 slope = m = _______ y -intercept ( , ) 8. 0 , 3/2 πŸπ’š = βˆ’π’™ +πŸ‘ π’š =βˆ’ 𝟏 𝟐 𝒙 + πŸ‘ 𝟐 πŸ“ πŸ‘ 5π‘₯βˆ’3𝑦 =βˆ’4 slope = m = _______ y -intercept ( , ) 9. 0 , 4/3 βˆ’πŸ‘π’š = βˆ’πŸ“π’™ βˆ’πŸ’ π’š = πŸ“ πŸ‘ 𝒙 + πŸ’ πŸ‘

9 U1B L2 Reviewing Linear Functions
EXAMPLE 10 Find the equation in slope-intercept form for the line with slope 𝟐 πŸ‘ and passes through the point (βˆ’πŸ‘, πŸ“) π’Ž= 𝟐 πŸ‘ Step 1: Solve for b using the point (βˆ’πŸ‘, πŸ“) π’š=π’Žπ’™+𝒃 πŸ“= 𝟐 πŸ‘ βˆ’πŸ‘ +𝒃 πŸ“=βˆ’πŸ+𝒃 (𝟎, πŸ•) (βˆ’πŸ‘, πŸ“) b = 7 Step 2: Find the equation π’š= 𝟐 πŸ‘ 𝒙+πŸ•

10 U1B L2 Reviewing Linear Functions
EXAMPLE 11 Find the equation in slope-intercept form for the line parallel to π’š = 𝟐 πŸ“ 𝒙 + 𝟐 and through the point (10, -1) Step 1: The slope of a parallel line will be 𝟐 πŸ“ Step 2: Solve for b using the point (𝟏𝟎, βˆ’πŸ) π’š=π’Žπ’™+𝒃 (𝟎, 𝟐) βˆ’πŸ= 𝟐 πŸ“ 𝟏𝟎 +𝒃 βˆ’πŸ=πŸ’+𝒃 𝒃 = βˆ’πŸ“ Step 3: Find the equation (𝟎, βˆ’πŸ“) π’š = 𝟐 πŸ“ 𝒙 – πŸ“

11 U1B L2 Reviewing Linear Functions
EXAMPLE 12 Write the equation for the line through the point (– 1 , 2) that is parallel to the line L: y = 3x – 4 Step 1: Slope of L is 3 so slope of any parallel line is also 3. Step 2: Find b. Step 3: The equation of the line parallel to L: π’š = πŸ‘π’™ – πŸ’ is π’š = πŸ‘π’™ + πŸ“ 𝟐 = πŸ‘(βˆ’πŸ) + 𝒃 𝟐 =βˆ’πŸ‘+𝒃 𝒃 = πŸ“ Step 4: Graph on your calculator to check your work. Use a square window. Y1 = 3x – 4 Y2 = 3x + 5 (0, 5) (0, – 4)

12 U1B L2 Reviewing Linear Functions
EXAMPLE 13 Write the equation for the line that is perpendicular to π’š = 𝟐 πŸ“ 𝒙 + 𝟐 and passes through the point (10, – 1 ) Step 1: The slope of a perpendicular line will be – πŸ“ 𝟐 negative reciprocal Step 2: Solve for b using the point (10, – 1) βˆ’πŸ = – πŸ“ 𝟐 (𝟏𝟎) + 𝒃 Step 3: The equation of the line β”΄ to π’š = 𝟐 πŸ“ 𝒙 + 𝟐 is π’š = – πŸ“ 𝟐 𝒙 + πŸπŸ’ βˆ’πŸ=βˆ’πŸπŸ“+𝒃 𝒃 = πŸπŸ’ Step 4: Graph on your calculator to check your work. Use a square window. Y1 = 𝟐 πŸ“ 𝒙 + 𝟐 Y2 = – πŸ“ 𝟐 x + 24 𝟐 𝟐 βˆ’πŸ“ πŸ“

13 U1B L2 Reviewing Linear Functions
EXAMPLE 14 Write the equation for the line through the point (– 1, 2) that is perpendicular to the line L: y = 3x – 4 Step 1: Slope of L is 3 so slope of any perpendicular line is βˆ’ 𝟏 πŸ‘ . Step 2: Find b. 𝟐 =βˆ’ 𝟏 πŸ‘ (βˆ’πŸ) + 𝒃 𝒃 = πŸ“ πŸ‘ 𝟐 = 𝟏 πŸ‘ + 𝒃 πŸ” πŸ‘ = 𝟏 πŸ‘ + 𝒃 Step 3: Find the equation of the line perpendicular to L: y = 3x – 4 π’š =βˆ’ 𝟏 πŸ‘ 𝒙 + πŸ“ πŸ‘ Step 4: Graph on your calculator to check your work. Use a square window. Y1 = 3x – 4 Y2 =βˆ’ 𝟏 πŸ‘ 𝒙 + πŸ“ πŸ‘

14 U1B L2 Reviewing Linear Functions
EXAMPLE 15 Find the equation in slope-intercept form for the line that passes through the points (7, βˆ’2) and (βˆ’5, 8). Step 1: Find the slope Step 2: Solve for b using either point π’Ž= βˆ’πŸβˆ’πŸ– πŸ•βˆ’(βˆ’πŸ“) = βˆ’πŸπŸŽ 𝟏𝟐 =βˆ’ πŸ“ πŸ” βˆ’πŸ =βˆ’ πŸ“ πŸ” (πŸ•) + 𝒃 πŸ– =βˆ’ πŸ“ πŸ” (βˆ’ πŸ“) + 𝒃 βˆ’ 𝟏𝟐 πŸ” =βˆ’ πŸ‘πŸ“ πŸ” + 𝒃 πŸ’πŸ– πŸ” = πŸπŸ“ πŸ” + 𝒃 𝒃 = πŸπŸ‘ πŸ” 𝒃 = πŸπŸ‘ πŸ” Step 3: Find the equation (– 5, 8) π’š =βˆ’ πŸ“ πŸ” 𝒙 + πŸπŸ‘ πŸ” (7, – 2)

15 U1B L2 Reviewing Linear Functions
EXAMPLE 16 Write the slope-intercept equation for the line through (– 2, –1) and (5, 4). Slope = m = βˆ’πŸβˆ’πŸ’ βˆ’πŸβˆ’πŸ“ = βˆ’πŸ“ βˆ’πŸ• = πŸ“ πŸ• β€“πŸ= πŸ“ πŸ• (β€“πŸ) + 𝒃 βˆ’πŸ• πŸ• = βˆ’πŸπŸŽ πŸ• + 𝒃 𝒃 = πŸ‘ πŸ• (5, 4) (– 2, – 1) Equation for the line is π’š = πŸ“ πŸ• 𝒙 + πŸ‘ πŸ•

16 U1B L2 Reviewing Linear Functions
Finish the 5 questions in Lesson #2


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