UNIT 1B LESSON 2 REVIEW OF LINEAR FUNCTIONS. Equations of Lines The horizontal line through the...
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Transcript of UNIT 1B LESSON 2 REVIEW OF LINEAR FUNCTIONS. Equations of Lines The horizontal line through the...
UNIT 1B LESSON 2
REVIEW OF LINEAR FUNCTIONS
Equations of Lines
The horizontal line through the point (2, 3) has equation
The vertical line through the point (2, 3) has equation
y = 3
x = 2
The vertical line through the point (a, b)has equation x = a since every x-coordinate
on the line has the same value a.
Similarly, the horizontal line through
(a, b) has equation y = b
(π ,π )
(π ,π )
(π ,π )(π ,βπ )
(βπ ,π ) (π ,π )(π ,π ) (π ,π )
Finding Equations of Vertical and Horizontal Lines
Vertical Line is x = β 3
Horizontal Line is y = 8
EXAMPLE 1 Write the equations of the vertical and horizontal lines through the point
(βπ ,π )
Y1 = 2x + 7
x Y = 2x + 7
y β intercept ( , )
π (βπ )+π=ππ (π )+π=π
π π
Slope y-intercept form y = mx + b
slope y-intercept (0, b)
EXAMPLE 2: Reviewing Slope-Intercept Form of Linear Functions
π=πππππππ
=ππβ ππ
ππβ ππ
π=πβπ
βπβπ=
βπβπ
=π
(π ,π)
(βπ ,π)
Unit 1B Lesson 2 Page 1EXAMPLESState the slopes and y-intercepts of the given linear functions.
y = 4x slope = m = _______ y -intercept ( , )3.
y = 3x β 5 slope = m = _______ y -intercept ( , )4.
)(xf 23
1x= slope = m = _______ y -intercept ( , )6.
xxf 12
1)(
slope = m = _______ y -intercept ( , )
6.
4
3
β
π (π )=ππ
βπππ
0 , 0
0
0 ,
0 ,
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.
y = β (A/B) x + C/B
By = β Ax + C
Ax + By = C
Analyzing and Graphing a General Linear Equation
Rearrange for y
Slope is
y-intercept is
Find the slope and y-intercept of the line
βπβπ
π=βπβπ
π+ππβπ
π=πππβπ
π
ππ
π
Example 7
Unit 1B Lesson 2 Page 1EXAMPLESState the slopes and y-intercepts of the given linear functions.
x + 2y = 3 slope = m = _______ y -intercept ( , )8.βππ
π π=βπ+π
0 , 3/2
π=βπππ+
ππ
slope = m = _______ y -intercept ( , )9.
βπ π=βπ πβπ
π=πππ+
ππ
ππ 0 , 4/3
π=ππ
(βπ )+π
EXAMPLE 10Find the equation in slope-intercept form for the line with slope and passes through the point
Step 1: Solve for b using the point
Step 2: Find the equation
(π ,π)(βπ ,π)
π=ππ+π
b = 7
π=βπ+π
π=ππ
βπ=ππ
(ππ )+π
EXAMPLE 11Find the equation in slope-intercept form for the line parallel to and through the point (10, -1)
Step 2: Solve for b using the point
Step 3: Find the equation
Step 1: The slope of a parallel line will be
π=βπβπ=π+π
π=ππ+π
(π ,βπ)
(π ,π)
EXAMPLE 12Write 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: is
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)
π=ππ=βπ+ππ=π(βπ)+π
EXAMPLE 13Write the equation for the line that is perpendicular to and passes through the point (10, β 1 )
Step 2: Solve for b using the point (10, β 1)
Step 3: The equation of the line β΄ to
is
Step 1: The slope of a perpendicular line will be negative reciprocal
Step 4: Graph on your calculator to check your work. Use a square window.
Y1 = Y2 = β x + 24
π=ππβπ=βππ+π
βππ
ππ
EXAMPLE 14Write 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 .
π=βππ
(βπ)+π
Step 3: Find the equation of the line perpendicular to L: y = 3x β 4
Step 4: Graph on your calculator to check your work. Use a square window.
Y1 = 3x β 4 Y2
Step 2: Find b.
π=πππ=
ππ
+πππ
=ππ
+π
βπ=βππ
(π)+π
EXAMPLE 15Find the equation in slope-intercept form for the line that passes through the points (7, 2) and (5, 8).
Step 1: Find the slope Step 2: Solve for b using either point
Step 3: Find the equation
π=βππ
(βπ)+ππ=βπβππβ(βπ)
=βππππ
=βππ
(7, β 2)
(β 5, 8)
π=πππ
βπππ
=βπππ
+π
π=πππ
πππ
=πππ
+π
EXAMPLE 16Write the slope-intercept equation for the line through (β 2, β1) and (5, 4).
Slope = m =
βπ=ππ
(βπ)+π
Equation for the line is
(5, 4)(β 2, β 1)
π=ππ
βππ
=βπππ
+π
Finish the 5 questions in Lesson #2