~ chapter 6 ~ solving equations algebra i lesson 6-1 rate of change & slope lesson 6-2...

~ Chapter 6 ~Solving

Equations

Algebra I

Lesson 6-1 Rate of Change & Slope

Lesson 6-2 Slope-Intercept Form

Lesson 6-3 Standard Form

Lesson 6-4 Point-Slope Form & Writing Linear Eq.

Lesson 6-5 Parallel & Perpendicular Lines

Lesson 6-6 Scatter Plots & Equations of Lines

Lesson 6-7 Graphing Absolute Value Equations

Chapter Review

Algebra I

Rate of Change &

Slope NotesLesson 6-1

Rate of change – change in the dependent variable or vertical changechange in the independent variable horizontal change

Find the rate of change…

120 - 60 5 - 1

60/4 = 15/1

Slope = vertical change = rise horizontal change run

Finding Slope using a graph…

Slope = 3 - 1 4 – (-1)

2/5

Choose two points

0 - 1000 180 - 60

-1000/ 120 =

- 25/3 or -8 1/3

Rate of Change &


Your turn…

Find the slope…

-1/6

Finding slope using points…

Slope = rise = y2 – y1 , where x2 – x1 ≠ 0

run x2 – x1

C (2,5) & D (4,7)

So… Slope = 7 - 5 4 – 2

2/2 = 1/1 = 1

Your turn… P(-1,4) & Q(3,-2)

Slope = -6/4 = -3/2

Horizontal lines have a slope of 0. Vertical lines have a slope that is undefined.

Rate of Change &


Find the slope…

Choose two points

Find the slope…

Choose two points

Rate of Change & Slope

HomeworkLesson 6-1

Homework – Practice 6-1 even

Slope-Intercept Form

Practice 6-1Lesson 6-2


NotesLesson 6-2

Writing Linear Equations

Linear equation – an equation whose graph is a line. (ex: y = 2x – 3)

y-intercept is the y coordinate of the point where the line crosses the y-axis.

If you know the slope of a line (use the slope formula) and the y-intercept… you can write a linear equation.

Write the linear equation…

Slope = 1 – 4 = -3/8 8 – 0So… y = -3/8x + 4

The slope-intercept form of a Linear equation

Find the slope and y-intercept of each equation…

y = -2x + 1 y = 7/6 x – ¾ y = -4/5 x

If you know the slope ( -3) and the y-intercept (4), write the linear equation…

y = -3 x + 4


NotesLesson 6-2

Writing an Equation from a graph…

Find the slope… (Choose 2 points on the line)

Slope = 1 – 2 = -1/-2 = 1/2 0 – 2

y = 1/2 x + 1

Graphing Equations

Step 1 – Plot the y-intercept point. (0, y-intercept)

Step 2 – Use the slope to plot the 2nd point.

Step 3 – Draw a line through the two points.

Graph y = 2x – 1

Graph y = 3/2 x - 2


HomeworkLesson 6-2

Homework ~ Practice 6-2 odd

Standard Form Practice 6-2

Lesson 6-3

Standard Form

NotesLesson 6-3

Standard form of a linear equation… Ax + By = C, where A, B, & C are real numbers and A & B are not both zero.

Finding the x- & y-intercepts

The x-intercept is the point on the line where y is zero, or (x-intercept, 0), so…

In 3x + 4y = 8, solve for x using the y value of 0

3x + 4(0) = 8

3x = 8

x = 8/3, or 2 2/3 so (8/3, 0) is where the line crosses the x-axis.

The y-intercept is the point on the line where x is zero, or (0, y-intercept), so…

In 3x + 4y = 8, solve for y using the x value of 0

3(0) + 4y = 8

4y = 8

y = 8/4 = 2, so (0, 2) is where the line crosses the y-axis.

Your turn… find the x- & y-intercepts for 4x – 9y = -12

Standard Form

NotesLesson 6-3

The x-intercept = -3 & the y-intercept = 4/3

Graphing using the x- & y-intercepts

Step 1 – Find the x- & y-intercepts

Step 2 – Plot the points (x-intercept, 0) & (0, y-intercept)

Step 3 – Connect the dots… Draw a line through the points.

Graph 5x + 2y = -10

Graphing horizontal & vertical lines…

y = -3 … write in standard form…

0x + 1y = -3, so for all values of x, y = -3

x = -4… write in standard form…

1x + 0y = -4, so for all values of y, x = -4

Transforming to standard form

Write -2/5x +1 in standard form (Ax + By = C)

Standard Form

NotesLesson 6-3

5y = 5(-2/5 x + 1)

5y = -2x + 5

2x + 5y = 5 (standard form)

You try… y= 2/3x + 5

-2x + 3y = 15 (standard form)

y = -4/5x – 7

4x + 5 y = -35 (standard form)

Standard Form

HomeworkLesson 6-3

Homework – Practice 6-3 odd

Point-Slope Form & Writing Linear Eq. Practice 6-3Lesson 6-4

Point-Slope Form & Writing Linear Eq.

NotesLesson 6-4

Using Point-Slope form… for a Linear Equation

The point-slope form of the equation of a non vertical line that passes through the point (x1, y1) with slope of m…

y – y1 = m (x – x1)

So, write the point slope form of the equation of a line through the point (3, 4) and with slope of 2

y – 4 = 2 (x – 3)

Your turn… slope of 2/5 and passes through the point (10, -8)

y – (-8) = 2/5 (x - 10) or y + 8 = 2/5 (x – 10)

Graphing using the point-slope form

Step 1 – Plot the point the equation shows.

Step 2 – Plot the next point using the slope.

Step 3 – Connect the dots.

Graph the equation y – 5 = ½ (x – 2)


NotesLesson 6-4

Graph y – 5 = -2/3(x + 2)

Using 2 Points to Write an Equation

Write equations for a line in point-slope form and in slope-intercept form.

Step 1 - Choose two points and find the slope

Step 2 – Use either point used to find the slope to write the equation in point-slope form.

Step 3 – Rewrite the equation from Step 2 in slope-intercept form.

Slope = -5 – 3 = -8/-3 = 8/3 -1 – 2

y – (-5) = 8/3 (x – (-1)) =

y + 5 = 8/3 (x + 1) (point-slope form)

y + 5 = 8/3x + 8/3

y = 8/3x – 2 1/3 (slope-intercept form)

Point-Slope Form & Writing Linear Eq. Homework

Lesson 6-4


Parallel & Perpendicular LinesPractice 6-4Lesson 6-5


NotesLesson 6-4

Writing an Equation Using a Table

Step 1 – Find the rate of change for consecutive ordered pairs. (Determine if the relationship is linear. Hint: each rate of change is same for each camparison)

Step 2 – Use the slope & a point to write an equation.

Rate of change for each is ½

So y – 6 = ½ (x - 3)

What about

In Summary ~

Slope-Intercept form: y = mx + b

Standard Form: Ax + By = C

Point-Slope Form: (y – y1) = m (x – x1)

x y

-1 4

3 6

5 7

11 10

x y

-11 -7

-1 -3

4 -1

19 5

Parallel & Perpendicular LinesNotesLesson 6-5

Slopes of parallel lines – nonvertical lines are parallel if they have the same slope and different y-intercepts. For example ~ y = ½ x + 3 & y = ½ x – 1 have the same slope of ½ and different y-intercepts so they are parallel.

Determine whether lines are parallel…

Step 1 – solve both equations for y (slope-intercept form)

Step 2 – compare slope and y-intercepts.

Determine if -6x + 8y = -24 & y = ¾ x – 7 are parallel.

8y = 6 x – 24

y = 6/8 x – 3

y = ¾ x – 3 & y = ¾ x – 7 parallel?

The lines are parallel. The equations have the same slope, ¾, & different y-intercepts.

Writing equations of parallel lines

Given a point and an equation, write the equation for the parallel line passing through the given point.

Parallel & Perpendicular LinesPractice 6-4Lesson 6-5


Step 1 – Identify the slope of the given line.

Step 2 – Use the given point to write the point-slope form equation and then convert to the slope-intercept form.

Write an equation for the line that contains (2, -6) and is parallel to y = 3x + 9

Slope = 3

y – (-6) = 3 (x – 2) (solve for y)

y + 6 = 3x – 6

y = 3x – 12

Slopes of Perpendicular lines – Two lines are perpendicular if the product of their slopes is -1. A vertical and a horizontal line are also perpendicular.

For example: slope is ¾ so the perpendicular line would have a slope of -4/3.

(Hint: The perpendicular line’s slope will be the negative reciprocal)

Slope = -2/5 Perpendicular slope = ?


Writing Equations for Perpendicular Lines

Given a point and an equation of a line, find the equation of a line perpendicular to the given line.

Step 1 – Identify the slope of the given line.

Step 2 – Convert the slope (negative reciprocal)

Step 3 – Write the point-slope form of the equation and convert to the slope-intercept form of the equation.

Write an equation of the line that contains (1, 8) and is perpendicular to y = ¾ x + 1.

Slope = ¾

Negative reciprocal = -4/3

y – 8 = -4/3(x – 1)

y – 8 = -4/3x + 4/3

y = -4/3x + 9 1/3

Parallel & Perpendicular LinesHomework

Lesson 6-5

Homework ~ Practice 6-5

#13-36

Scatter Plots & Equations of Lines


Graphing Absolute Value EquationsNotesLesson 6-7

Absolute value equation – V-shaped graph pointing upward or downward.

Translation – shift of a graph horizontally, vertically, or both. (slide)

Vertical Translations y = |x|

So, what would y = |x| + 2 look like?

What about y = |x| - 3?

So, the graph of y = |x| + k is a translation of y = |x|. If k is positive then the graph translates up k units, and if k is negative then the graph translates down k units.

To show a vertical translation… graph y = |x| and then graph the translation y = |x| + k


Writing an Absolute Value Equation

Write an equation for each translation of y = |x|…

8 units down 6 units up 2 units up 5 units down

y = |x| - 8 y = |x| + 6 y = |x| + 2 y = |x| - 5

Horizontal Translations

y = |x + h| is a translation of y = |x|. If h is positive then the graph translates h units to the left. If h is negative then the graph translates h units to the right.

Graph the equation by translating y = |x| for y = |x - 2|

What about y = |x + 2|?

What about y = |x - 4|?


Writing an Absolute Value Equation

Write an equation for each translation of y = |x|…

8 units left 6 units right 5 unit right 7 units left

y = |x + 8| y = |x -6| y = |x - 5| y = |x + 7|

Graphing Absolute Value Equations

HomeworkLesson 6-7


~ Chapter 6 ~Chapter Review

Algebra I Algebra I

~ chapter 6 ~ solving equations algebra i lesson 6-1 rate of change & slope lesson 6-2...

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