Download presentation

Presentation is loading. Please wait.

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. 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

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

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 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

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 = β Ax + C y = β (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₯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

8
**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

9
**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

10
**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

11
**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)

12
**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

13
**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

14
**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)

15
**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

16
**U1B L2 Reviewing Linear Functions**

Finish the 5 questions in Lesson #2

Similar presentations

Presentation is loading. Please wait....

OK

Rates of Change (Slope)

Rates of Change (Slope)

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on natural numbers math Ppt on polynomials in maths class Ppt on different solid figures and nets Ppt on dairy milk silk Ppt on working of nuclear power plant in india Ppt on circles theorems for class 9 Ppt on c++ basic programing Ppt on related party transactions under companies act 2013 Ppt on waxes biology Download ppt on cultural heritage of india