Warm-Up Khan Questions Review New Topic Today:
Warm-UpKhan Questions
Review New Topic
Today:
Warm-Up
Your double agent is lost in the woods. You have located her on the grid at (-2,-3). The enemy is closing in fast, and the only two Safe houses are located in the direction of a -2/3 slope from her present location.
Help your agent narrowly escape death by sending her the coordinates of the two safe houses.
Warm-Up 1. What is 250% of 18? 2. 24 is reduced by 25%; what number could you multiply 24 by to get the answer? 3. Name three solutions to the equation y = -4x+1 4. Find the x and y intercepts, and the slope, then graph the equation 2x = 3y -6
5. Find the slope of the line containing the following points: (0,5) (3,7) (6,9) (9,11)
6. The slope of the line between (2,10) and (x,4) is -3. Find the value of x.
7. 7|x−10|−9 = 54
Views of a Function
Views of a Function
Domain & Range
Domain & Range
Identifying Linear RelationshipsWhat is y when x is
4?
Identifying Linear RelationshipsWhat is x when y is 4?
Identifying Linear Relationships
How does y change as x increases?
Finding the Equation of a LineGraph the following equation: y = -3x - 7
Monday's Topic: Graphing Equations with
one variableAs you know, every point on a coordinate plane is the intersection of two variables, the independent x, and the dependent y.
What if, however, your equation only has an independent x variable? 2x - 2 = -4
These equations can still be solved by finding the x intercept, since we know that at this point, y = 0
How is it Done?
Task: Graph the equation 2x - 2 = - 4 1. First, set the equation equal to zero: 2x + 2 = 0 2. Replace 0 with f(x) 3. Make a table 4. Graph the ordered pairsGraph the Equation
** Graph 5x + 2 = 7
Today's Topic: Using the Slope-Intercept Form
Finding the Equation of a Line
11-3
Using Slopes and Intercepts
Course 3
Learn to use slopes and intercepts to graph linear equations.
Course 3
11-3 Using Slopes and Intercepts
As you watch the video, take notes on your handout.
Insert Lesson Title Here
Course 3
11-3
One way to graph a linear equation easily is by finding the x-intercept and the y-intercept.
The x-intercept is the value of x where the line crosses the x-axis (y = 0).
The y-intercept is the value of y where the line crosses the y-axis (x = 0).
Course 3
11-3 Using Intercepts
Find the x-intercept and y-intercept of the line 4x – 3y = 12. Use the intercepts to graph the equation.
Example 1
Find the x-intercept (y = 0).
4x – 3y = 12
4x – 3(0) = 12
4x = 124x4
124=
x = 3The x-intercept is 3.
Course 3
11-3 Using Intercepts
Example 1 Continued
Find the y-intercept (x = 0).
4x – 3y = 12
4(0) – 3y = 12
–3y = 12
-3y-3
12-3 =
y = –4
The y-intercept is –4.
Course 3
11-3 Using Intercepts
4x – 3y = 12
Crosses the x-axis at the point (3, 0)
Crosses the y-axis at the point (0, –4)
Course 3
11-3 Using Intercepts
Find the x-intercept and y-intercept of the line 8x – 6y = 48. Use the intercepts to graph the equation.
Try This
Find the x-intercept (y = 0).
8x – 6y = 48
8x – 6(0) = 48
8x = 488x8
488=
x = 6The x-intercept is 6 so the point is (6, 0).
Course 3
11-3 Using Intercepts
Try This
Find the y-intercept (x = 0).
8x – 6y = 48
8(0) – 6y = 48
–6y = 48
-6y-6
48-6 =
y = –8
The y-intercept is –6 so the point is (0, -8).
Course 3
11-3 Using Intercepts
Try This: Example 1 Continued
The graph of 8x – 6y = 48 is the line that crosses the x-axis at the point (6, 0) and the y-axis at the point (0, –8).
Course 3
11-3 Using Slopes and Intercepts
In an equation written in slope-intercept form, y = mx + b, m is the slope and b is the y-intercept.
y = mx + b
Slope y-intercept
Course 3
11-3 Using Slopes and Intercepts
For an equation such as y = x – 6, write it as y = x + (–6) to read the y-intercept, –6. The point would be (0,-6)
Helpful Hint
Course 3
11-3 Using Slopes and Intercepts
Using the Slope-Intercept Form
1. Isolate the y so that you equation is in y = mx+b form.
2. Slope will always be “m” (the number in front of x).
3. The y-intercept will always be “b” (the number by itself). Write this point as (0,b)
Example 2
Write each equation in slope-intercept form, and then find the slope and y-intercept.
A. 2x + y = 32x + y = 3–2x –2x Subtract 2x from both sides.
y = 3 – 2x
y = –2x + 3 The equation is in slope-intercept form.
m = –2 b = 3The slope of the line is –2, and the y-intercept is 3.
Course 3
11-3 Using Slopes and Intercepts
More Examples
B. 5y = 3x
5y = 3x
Divide both sides by 5 to solve for y.
The equation is in slope-intercept form.
b = 0
= x35
5y5
y = x + 035
m =35
The slope of the line is , and they-intercept is 0.
35
Course 3
11-3 Using Slopes and Intercepts
More Examples
C. 4x + 3y = 9
4x + 3y = 9Subtract 4x from both sides.
b = 3
y =- x + 343
m =- 43
The slope of the line 4x+ 3y = 9 is – , and the y-intercept is 3.4
3
–4x –4x
3y = –4x + 9
= + –4x 3
3y3
93 Divide both sides by 3.
The equation is in slope-intercept form.
Course 3
11-3 Using Slopes and Intercepts
Try This
Write each equation in slope-intercept form, and then find the slope and y-intercept.
A. 4x + y = 4–4x –4x Subtract 4x from both sides.
y = 4 – 4xRewrite to match slope-intercept form.
y = –4x + 4 The equation is in slope-intercept form.
m = –4 b = 4The slope of the line 4x + y = 4 is –4, and the y-intercept is 4.
Course 3
11-3 Using Slopes and Intercepts
Try This
B. 7y = 2x
7y = 2x
Divide both sides by 7 to solve for y.
The equation is in slope-intercept form.
b = 0
= x27
7y7
y = x + 027
m =27
The slope of the line 7y = 2x is , and they-intercept is 0.
27
Course 3
11-3 Using Slopes and Intercepts
Try This
C. 5x + 4y = 8
5x + 4y = 8Subtract 5x from both sides.
Rewrite to match slope-intercept form.
b = 2
y =- x + 254
The slope of the line 5x + 4y = 8 is – , and the y-intercept is 2.5
4
–5x –5x
4y = 8 – 5x
5x + 4y = 8
= + –5x 4
4y4
84 Divide both sides by 4.
The equation is in slope-intercept form.
m =- 54
Course 3
11-3 Using Slopes and Intercepts
Additional Example 3: Entertainment Application
A video club charges $8 to join, and $1.25 for each DVD that is rented. The linear equation y = 1.25x + 8 represents the amount of money y spent after renting x DVDs. Graph the equation by first identifying the slope and y-intercept.
y = 1.25x + 8The equation is in slope-intercept form.
b = 8m =1.25
Course 3
11-3 Using Slopes and Intercepts
Additional Example 3 Continued
The slope of the line is 1.25, and the y-intercept is 8. The line crosses the y-axis at the point (0, 8) and moves up 1.25 units for every 1 unit it moves to the right.
Course 3
11-3 Using Slopes and Intercepts
Try This: Example 3
A salesperson receives a weekly salary of $500 plus a commission of 5% for each sale. Total weekly pay is given by the equation S = 0.05c + 500. Graph the equation using the slope and y-intercept.
y = 0.05x + 500 The equation is in slope-intercept form.
b = 500m =0.05
Course 3
11-3 Using Slopes and Intercepts
Try This: Example 3 Continued
The slope of the line is 0.05, and the y-intercept is 500. The line crosses the y-axis at the point (0, 500) and moves up 0.05 units for every 1 unit it moves to the right.
x
y
500
1000
1500
2000
10,0005000 15,000
Course 3
11-3 Using Slopes and Intercepts
Additional Example 4: Writing Slope-Intercept Form
Write the equation of the line that passes through (3, –4) and (–1, 4) in slope-intercept form.
Find the slope.
The slope is –2.
Choose either point and substitute it along with the slope into the slope-intercept form.
y = mx + b
4 = –2(–1) + b
4 = 2 + b
Substitute –1 for x, 4 for y, and –2 for m.
Simplify.
4 – (–4) –1 – 3
=y2 – y1
x2 – x1
8–4= = –2
Course 3
11-3 Using Slopes and Intercepts
Additional Example 4 Continued
Solve for b.
Subtract 2 from both sides.
Write the equation of the line, using –2 for m and 2 for b.
4 = 2 + b–2 –2
2 = b
y = –2x + 2
Course 3
11-3 Using Slopes and Intercepts
Try This: Example 4
Write the equation of the line that passes through (1, 2) and (2, 6) in slope-intercept form.
Find the slope.
The slope is 4.
Choose either point and substitute it along with the slope into the slope-intercept form.
y = mx + b
2 = 4(1) + b
2 = 4 + b
Substitute 1 for x, 2 for y, and 4 for m.
Simplify.
6 – 2 2 – 1
=y2 – y1
x2 – x1
4 1= = 4
Course 3
11-3 Using Slopes and Intercepts
Try This: Example 4 Continued
Solve for b.
Subtract 4 from both sides.
Write the equation of the line, using 4 for m and –2 for b.
2 = 4 + b–4 –4
–2 = b
y = 4x – 2
Course 3
11-3 Using Slopes and Intercepts
11-4 Point-Slope Form
Course 3
Warm Up
Problem of the Day
Lesson Presentation
A-CED.2: Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
Warm UpWrite the equation of the line that passes through each pair of points in slope-intercept form.
1. (0, –3) and (2, –3)
2. (5, –3) and (5, 1)
3. (–6, 0) and (0, –2)
4. (4, 6) and (–2, 0)
y = –3
x = 5
Course 3
11-4 Point-Slope Form
y = x + 2
y = – x – 213
Problem of the Day
Without using equations for horizontal or vertical lines, write the equations of four lines that form a square.
Possible answer: y = x + 2, y = x – 2, y = –x + 2, y = –x – 2
Course 3
11-4 Point-Slope Form
Learn to find the equation of a line given one point and the slope.
Course 3
11-4 Point-Slope Form
Vocabulary
point-slope form
Insert Lesson Title Here
Course 3
11-4 Point-Slope Form
Point on the line(x1, y1)
Point-slope formy – y1 = m (x – x1)
slope
The point-slope of an equation of a line with slope m passing through (x1, y1) is y – y1 = m(x – x1).
Course 3
11-4 Point-Slope Form
Use the point-slope form of each equation to identify a point the line passes through and the slope of the line.
A. y – 7 = 3(x – 4)
Additional Example 1: Using Point-Slope Form to Identify Information About a Line
y – y1 = m(x – x1)
y – 7 = 3(x – 4)
m = 3
(x1, y1) = (4, 7)
The line defined by y – 7 = 3(x – 4) has slope 3, and passes through the point (4, 7).
The equation is in point-slope form. Read the value of m from the equation. Read the point from the equation.
Course 3
11-4 Point-Slope Form
B. y – 1 = (x + 6)
Additional Example 1B: Using Point-Slope Form to Identify Information About a Line
y – y1 = m(x – x1)
(x1, y1) = (–6, 1)
Rewrite using subtraction instead of addition.
13
13
y – 1 = (x + 6)
y – 1 = [x – (–6)]13
m =13
The line defined by y – 1 = (x + 6) has slope , and
passes through the point (–6, 1).
13
13
Course 3
11-4 Point-Slope Form
Use the point-slope form of each equation to identify a point the line passes through and the slope of the line.
A. y – 5 = 2 (x – 2)
Try This: Example 1
y – y1 = m(x – x1)
y – 5 = 2(x – 2)
m = 2
(x1, y1) = (2, 5)
The line defined by y – 5 = 2(x – 2) has slope 2, and passes through the point (2, 5).
The equation is in point-slope form. Read the value of m from the equation. Read the point from the equation.
Course 3
11-4 Point-Slope Form
B. y – 2 = (x + 3)Try This: Example 1B
23
(x1, y1) = (–3, 2)
Rewrite using subtraction instead of addition.
23
y – 2 = (x + 3)
y – 2 = [x – (–3)]23
m =23
The line defined by y – 2 = (x + 3) has slope , and
passes through the point (–3, 2).
23
23
y – y1 = m(x – x1)
Course 3
11-4 Point-Slope Form
Write the point-slope form of the equation with the given slope that passes through the indicated point.
A. the line with slope 4 passing through (5, -2)
Additional Example 2: Writing the Point-Slope Form of an Equation
y – y1 = m(x – x1)
The equation of the line with slope 4 that passes through (5, –2) in point-slope form is y + 2 = 4(x – 5).
Substitute 5 for x1, –2 for y1, and 4 for m.
[y – (–2)] = 4(x – 5)
y + 2 = 4(x – 5)
Course 3
11-4 Point-Slope Form
B. the line with slope –5 passing through (–3, 7)
Additional Example 2: Writing the Point-Slope Form of an Equation
y – y1 = m(x – x1)
The equation of the line with slope –5 that passes through (–3, 7) in point-slope form is y – 7 = –5(x + 3).
Substitute –3 for x1, 7 for y1, and –5 for m.
y – 7 = -5[x – (–3)]
y – 7 = –5(x + 3)
Course 3
11-4 Point-Slope Form
Write the point-slope form of the equation with the given slope that passes through the indicated point.
A. the line with slope 2 passing through (2, –2)
Try This: Example 2A
y – y1 = m(x – x1)
The equation of the line with slope 2 that passes through (2, –2) in point-slope form is y + 2 = 2(x – 2).
Substitute 2 for x1, –2 for y1, and 2 for m.
[y – (–2)] = 2(x – 2)
y + 2 = 2(x – 2)
Course 3
11-4 Point-Slope Form
B. the line with slope -4 passing through (-2, 5)
Try This: Example 2B
y – y1 = m(x – x1)
The equation of the line with slope –4 that passes through (–2, 5) in point-slope form is y – 5 = –4(x + 2).
Substitute –2 for x1, 5 for y1, and –4 for m.
y – 5 = –4[x – (–2)]
y – 5 = –4(x + 2)
Course 3
11-4 Point-Slope Form
A roller coaster starts by ascending 20 feet for every 30 feet it moves forward. The coaster starts at a point 18 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 150 feet forward. Assume that the roller coaster travels in a straight line for the first 150 feet.
Additional Example 3: Entertainment Application
As x increases by 30, y increases by 20, so the slope
of the line is or . The line passes through the point (0, 18).
2030
23
Course 3
11-4 Point-Slope Form
Additional Example 3 Continued
y – y1 = m(x – x1) Substitute 0 for x1, 18 for y1,
and for m.23
The equation of the line the roller coaster travels along, in point-slope form, is y – 18 = x. Substitute 150 for x to find the value of y.
23
y – 18 = (150)23
y – 18 = 100
y – 18 = (x – 0)23
y = 118
The value of y is 118, so the roller coaster will be at a height of 118 feet after traveling 150 feet forward.
Course 3
11-4 Point-Slope Form
Try This: Example 3A roller coaster starts by ascending 15 feet for every 45 feet it moves forward. The coaster starts at a point 15 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 300 feet forward. Assume that the roller coaster travels in a straight line for the first 300 feet.
As x increases by 45, y increases by 15, so the slope
of the line is or . The line passes through the point (0, 15).
1545
13
Course 3
11-4 Point-Slope Form
Try This: Example 3 Continued
y – y1 = m(x – x1) Substitute 0 for x1, 15 for y1,
and for m.13
The equation of the line the roller coaster travels along, in point-slope form, is y – 15 = x. Substitute 300 for x to find the value of y.
13
y – 15 = (300)13
y – 15 = 100
y – 15 = (x – 0)13
y = 115
The value of y is 115, so the roller coaster will be at a height of 115 feet after traveling 300 feet forward.
Course 3
11-4 Point-Slope Form