# ~ Chapter 6 ~ Algebra I Algebra I Solving Equations

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~ Chapter 6 ~ Algebra I Algebra I Solving Equations
Lesson Rate of Change & Slope Lesson Slope-Intercept Form Lesson Standard Form Lesson Point-Slope Form & Writing Linear Eq. Lesson Parallel & Perpendicular Lines Lesson Scatter Plots & Equations of Lines Lesson Graphing Absolute Value Equations Chapter Review

Rate of Change & Slope Notes Lesson 6-1
Rate of change – change in the dependent variable or vertical change change in the independent variable horizontal change Find the rate of change… 5 - 1 60/4 = 15/1 Slope = vertical change = rise horizontal change run Finding Slope using a graph… Slope = 4 – (-1) 2/5 Choose two points -1000/ 120 = - 25/3 or -8 1/3

Rate of Change & Slope Notes Lesson 6-1 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 = 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 & Slope Notes Lesson 6-1 Find the slope…
Choose two points Find the slope… Choose two points

Lesson 6-1 Homework Homework – Practice 6-1 even
Rate of Change & Slope Homework Homework – Practice 6-1 even

Slope-Intercept Form Lesson 6-2 Practice 6-1

Notes Lesson 6-2 Slope-Intercept Form 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 – 0 So… y = -3/8x + 4 The slope-intercept form of a Linear equation Find the slope and y-intercept of each equation… y = -2x 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

Notes Lesson 6-2 Slope-Intercept Form
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

Lesson 6-2 Slope-Intercept Form Homework Homework ~ Practice 6-2 odd

Standard Form Lesson 6-3 Practice 6-2

Standard Form Lesson 6-3 Practice 6-2

Notes Standard Form Lesson 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

Lesson 6-3 Notes Standard Form
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)

Lesson 6-3 Notes Standard Form 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)

Lesson 6-3 Standard Form Homework Homework – Practice 6-3 odd

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

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

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

Point-Slope Form & Writing Linear Eq.
Lesson 6-4 Notes 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.
Lesson 6-4 Point-Slope Form & Writing Linear Eq. Homework Homework – Practice 6-4 odd

Parallel & Perpendicular Lines
Lesson 6-5 Practice 6-4

Parallel & Perpendicular Lines
Lesson 6-5 Practice 6-4

Point-Slope Form & Writing Linear Eq.
Lesson 6-4 Notes 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 19 5

Parallel & Perpendicular Lines
Lesson 6-5 Notes 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 Lines
Lesson 6-5 Practice 6-4

Parallel & Perpendicular Lines
Lesson 6-5 Notes 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 = ?

Parallel & Perpendicular Lines
Lesson 6-5 Notes 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 /3

Parallel & Perpendicular Lines
Lesson 6-5 Homework Homework ~ Practice 6-5 #13-36

Scatter Plots & Equations of Lines
Lesson 6-6 Practice 6-5

Scatter Plots & Equations of Lines
Lesson 6-6 Practice 6-5

Graphing Absolute Value Equations
Lesson 6-7 Notes 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

Graphing Absolute Value Equations
Lesson 6-7 Notes Writing an Absolute Value Equation Write an equation for each translation of y = |x|… 8 units down units up units up units down y = |x| y = |x| y = |x| 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|?

Graphing Absolute Value Equations
Lesson 6-7 Notes Writing an Absolute Value Equation Write an equation for each translation of y = |x|… 8 units left units right unit right units left y = |x + 8| y = |x -6| y = |x - 5| y = |x + 7|

Graphing Absolute Value Equations
Lesson 6-7 Graphing Absolute Value Equations Homework Homework – Practice 6-7 odd

Graphing Absolute Value Equations
Lesson 6-7 Practice 6-6

Graphing Absolute Value Equations
Lesson 6-7 Graphing Absolute Value Equations Practice 6-7

Graphing Absolute Value Equations
Lesson 6-7 Graphing Absolute Value Equations Practice 6-7

~ Chapter 6 ~ Algebra I Algebra I Chapter Review

~ Chapter 6 ~ Algebra I Algebra I Chapter Review

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