Dec.18

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

http://www.brainpop.com/math/algebra/slopeandintercept

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


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


4x – 3y = 12

Crosses the x-axis at the point (3, 0)

Crosses the y-axis at the point (0, –4)

Course 3


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


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


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


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


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


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


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


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.


Course 3


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


Try This

B. 7y = 2x

7y = 2x

Divide both sides by 7 to solve for y.


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


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.


m =- 54

Course 3


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


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


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



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


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



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


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



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


11-4 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


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


Learn to find the equation of a line given one point and the slope.

Course 3


Vocabulary

point-slope form

Insert Lesson Title Here

Course 3


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


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


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


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


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


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


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


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


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


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



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


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



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


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Dec.18