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Today: News & Notes Warm-Up Review: Graphing Inequalities Khan Academy Practice Questions Introduce: parallel & Perpendicular Lines
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Page 1: January 14

Today:News & Notes

Warm-UpReview: Graphing Inequalities

Khan Academy Practice QuestionsIntroduce: parallel & Perpendicular Lines

Page 2: January 14

News & Notes:

1. very nice work on Khan Academy this past week and weekend. I am sincerely impressed.

Day: total minutes: hours:

Thursday 829 13.82 Friday 1158 19.30Saturday 1154 19.223-day totals: 3141 52.35

2. Test this week (thurs. or Fri.) on graphing inequalities, parallel and perpendicular lines. what you do, (or don't do) everyday affects your grade. Be responsible, you'll feel good, really.

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Warm-Up:

2. From the table, determine the function, fill in the missing values, and write the equation solving for y (y = )

1.

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Warm-Up:3. From the table, determine the function, fill in the missing values, and write the equation solving for y (y = )

4.

(7x + 2) - (3x - 5)

4.

(3x + 2) (5x + 4) 5.

6.

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Warm-Up:7. 6a²c² + 2ac² + 10ab²c3 Find the GCF: 8. Simplify:

9.

10.

a. What's wrong with this picture?

b. 40 ounces cost $5.20. What's the unit price?

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Khan Academy Practice

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Khan Academy Practice

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Khan Academy Practice

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Khan Academy Practice

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Parallel & Perpendicular Lines

Two lines are

parallel

The lines never

intersect

Slopes are

equal

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parallel lines have the same slopes so find the slope of the given line and the slope you want for a parallel line will be the same.

Perpendicular lines have negative reciprocal slopes. take the slope of the given line, switch the numerator and denominator and change the sign for the slope of its perpendicular line.

3

1m

3

1m

3

1m

31

3m

Parallel & Perpendicular Lines

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(2, 4)

Let's find the equation of a line parallel to y = - x that passes through the point (2, 4)y = - x

What is the slope of the first line, y = - x ?

This is in slope intercept form so y = mx + b which means the slope is –1.

1

use the point-slope formula to find the equation of the line

Parallel & Perpendicular Lines

y -y1= -(x - x1) y -4 = -(x - 2)

y = -x + 6

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(2, 4)

Let's find the equation of a line perpendicular to y = - x that passes thru the point (2, 4)y = - x

The slope of the first line is still –1.

The slope of a line perpendicular is the negative reciporical so take –1 and "flip" it over and make it negative.

1

1

1

1

So the slope of a perpendicular line is 1 and it passes through (2, 4).

Parallel & Perpendicular Lines

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Parallel & Perpendicular Lines

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Parallel & Perpendicular Lines

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Class Work:

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