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**U1B L2 Reviewing Linear Functions**

UNIT 1B LESSON 2 REVIEW OF LINEAR FUNCTIONS

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**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. π,π βπ,π π,π π,π π,π π,π Similarly, the horizontal line through (a, b) has equation y = b π,π The horizontal line through the point (2, 3) has equation π,βπ y = 3 The vertical line through the point (2, 3) has equation x = 2

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**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 βπ, π Horizontal Line is y = 8 βπ, π Vertical Line is x = β 3

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**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 β3 π βπ +π=π π π +π=π y β intercept ( , ) π π (π,π) π= ππππ πππ = π π β π π π π β π π π= πβπ βπβπ = βπ βπ =π (βπ,π)

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**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 β slope = m = _______ y -intercept ( , ) 4. 3 0 , βπ = slope = m = _______ y -intercept ( , ) 6. β
0 , βπ slope = m = _______ y -intercept ( , ) 6. β π π 0 , Β½ π π = π π β π π π

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**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 = β Ax + C y = β (A/B) x + C/B

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**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₯β3π¦=15 Rearrange for y βππ = βππ + ππ βπ βπ π = βπ βπ π+ ππ βπ π π= π π πβπ π π π y-intercept is (π, βπ) Slope is π π

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**U1B L2 Reviewing Linear Functions**

Unit 1B Lesson 2 Page 1 EXAMPLES State the slopes and y-intercepts of the given linear functions. β π π x + 2y = 3 slope = m = _______ y -intercept ( , ) 8. 0 , 3/2 ππ = βπ +π π =β π π π + π π π π 5π₯β3π¦ =β4 slope = m = _______ y -intercept ( , ) 9. 0 , 4/3 βππ = βππ βπ π = π π π + π π

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**U1B L2 Reviewing Linear Functions**

EXAMPLE 10 Find the equation in slope-intercept form for the line with slope π π and passes through the point (βπ, π) π= π π Step 1: Solve for b using the point (βπ, π) π=ππ+π π= π π βπ +π π=βπ+π (π, π) (βπ, π) b = 7 Step 2: Find the equation π= π π π+π

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**U1B L2 Reviewing Linear Functions**

EXAMPLE 11 Find the equation in slope-intercept form for the line parallel to π = π π π + π and through the point (10, -1) Step 1: The slope of a parallel line will be π π Step 2: Solve for b using the point (ππ, βπ) π=ππ+π (π, π) βπ= π π ππ +π βπ=π+π π = βπ Step 3: Find the equation (π, βπ) π = π π π β π

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**U1B L2 Reviewing Linear Functions**

EXAMPLE 12 Write the equation for the line through the point (β 1 , 2) that is parallel to the line L: y = 3x β 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 = ππ β π is π = ππ + π π = π(βπ) + π π =βπ+π π = π Step 4: Graph on your calculator to check your work. Use a square window. Y1 = 3x β 4 Y2 = 3x + 5 (0, 5) (0, β 4)

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**U1B L2 Reviewing Linear Functions**

EXAMPLE 13 Write the equation for the line that is perpendicular to π = π π π + π and passes through the point (10, β 1 ) Step 1: The slope of a perpendicular line will be β π π negative reciprocal Step 2: Solve for b using the point (10, β 1) βπ = β π π (ππ) + π Step 3: The equation of the line β΄ to π = π π π + π is π = β π π π + ππ βπ=βππ+π π = ππ Step 4: Graph on your calculator to check your work. Use a square window. Y1 = π π π + π Y2 = β π π x + 24 π π βπ π

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**U1B L2 Reviewing Linear Functions**

EXAMPLE 14 Write the equation for the line through the point (β 1, 2) that is perpendicular to the line L: y = 3x β 4 Step 1: Slope of L is 3 so slope of any perpendicular line is β π π . Step 2: Find b. π =β π π (βπ) + π π = π π π = π π + π π π = π π + π Step 3: Find the equation of the line perpendicular to L: y = 3x β 4 π =β π π π + π π Step 4: Graph on your calculator to check your work. Use a square window. Y1 = 3x β 4 Y2 =β π π π + π π

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**U1B L2 Reviewing Linear Functions**

EXAMPLE 15 Find the equation in slope-intercept form for the line that passes through the points (7, β2) and (β5, 8). Step 1: Find the slope Step 2: Solve for b using either point π= βπβπ πβ(βπ) = βππ ππ =β π π βπ =β π π (π) + π π =β π π (β π) + π β ππ π =β ππ π + π ππ π = ππ π + π π = ππ π π = ππ π Step 3: Find the equation (β 5, 8) π =β π π π + ππ π (7, β 2)

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**U1B L2 Reviewing Linear Functions**

EXAMPLE 16 Write the slope-intercept equation for the line through (β 2, β1) and (5, 4). Slope = m = βπβπ βπβπ = βπ βπ = π π βπ= π π (βπ) + π βπ π = βππ π + π π = π π (5, 4) (β 2, β 1) Equation for the line is π = π π π + π π

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**U1B L2 Reviewing Linear Functions**

Finish the 5 questions in Lesson #2

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Graph a linear equation Graph: 2x β 3y = -12 Solve for y so the equation looks like y = mx + b - 3y = -2x β 12 Subtract 2x to both sides. y = x + 4 Divide.

Graph a linear equation Graph: 2x β 3y = -12 Solve for y so the equation looks like y = mx + b - 3y = -2x β 12 Subtract 2x to both sides. y = x + 4 Divide.

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