after today, the next four class periods are: review for quiz 2 quiz 2 (on sections 3.1-3.5) review...
TRANSCRIPT
![Page 1: After today, the next four class periods are: Review for Quiz 2 Quiz 2 (on sections 3.1-3.5) Review for Test 1 Test 1 (Chapters 1, 2, 3)](https://reader035.vdocuments.mx/reader035/viewer/2022062407/56649d0a5503460f949dce44/html5/thumbnails/1.jpg)
After today, the next four class periods are:
• Review for Quiz 2 • Quiz 2 (on sections 3.1-3.5)• Review for Test 1• Test 1 (Chapters 1, 2, 3)
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Any questions on the Section 3.4 homework?
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Please
CLOSE
YOUR LAPTOPS,and turn off and put away your
cell phones,
and get out your note-taking materials.
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Section 3.5Equations of Lines
Recall that the slope-intercept form of a line is
y = mx + b , where the line has a slope of m
and has a y-intercept of (0, b).
• If we know the slope and y-intercept of a line, we can substitute into this form to get an equation for the line.
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Example
• By substituting the appropriate values into the slope-intercept form, we get
y = -3x – .5
1
If you prefer to not use fractions in your final answer, multiply by 5 to get 15x + 5y = -1. (This is the way “standard form” is usually written, w/o fractions.)
Note: If you’re asked to write the equation in standard form, the answer could be converted to 3x + y = .
5
1
5
1
Find an equation of a line with slope of -3 and y-intercept of (0, ).
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Example
Graph y = x – 2.5
4
run
riseslope = , which in this case is 5
4
• We can use the slope-intercept form to help us graph the equation.
• We know that the y-intercept is (0, -2), which gives us one point for the line.
• We can also use the definition of slope to help us get another point.
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Example (cont.)
x
y
(0, -2)
First we graph the y-intercept.
Then we use the slope of -4/5 to find another point.
Move down 4 and to the right 5.
4 units down
5 units right
This gives us the new point (5, -6).
(5, -6)
Now draw the line.
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The slope-intercept form uses, specifically, the y-intercept in the equation.
The point-slope form allows you to use ANY point,
together with the slope, to form the equation of the line.
Point-slope formula for linear equations:
)( 11 xxmyy
Where m is the slope, and
(x1, y1) is a point on the line
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ATTENTION!Pay special attention to this next
slide and the examples that follow, because SEVERAL problems on both Quiz 2
and Test 1 will use this formula!
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Find an equation of a line with slope –2, through the point (-11,-12). Write the final equation in slope-intercept form.
(Note: it’s always a good idea to graph the line first. This will help you see if your equation makes sense, which is especially helpful on quizzes and tests.
• Solution: Substitute the slope and point into the point-slope form of the linear equation:
y – (-12) = -2(x – (-11)) y + 12 = -2x – 22 (use distributive property)
y = -2x - 34 (subtract 12 from both sides)
So the slope is -2, and the y-intercept is (0,-34)
Example )( 11 xxmyy
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Problem from today’s homework:
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Problem from today’s homework:
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Find the equation of the line through (-4,0) and (6,-1). Write the equation in standard form.
• First find the slope.
10
1
)4(6
01
12
12
xx
yym
Example
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Example (cont.)
• Now substitute the slope and one of the points into the point-slope form of an equation.
))4((10
10 xy
)4(110 xy (clear fractions by multiplying both sides by 10)
410 xy (use distributive property)
410 yx (add x to both sides)
NOTE: In slope-intercept form, this would be y = - 1 x - 2 10 5
)( 11 xxmyy
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Problem from today’s homework:
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Example
Find the equation of the line passing through points (2, 5) and (-4, 3). Write the equation using function notation.
First, calculate the slope:
3
1
6
2
)4(2
35
12
12
xx
yym
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Example (cont.)
))4((3
13 xy
3
4
3
13 xy
3
13
3
13
3
4
3
1 xxy
3
13
3
1)( xxf
Now enter the slope and one of the points (either one will work) into the point-slope equation:
(Looks just like slope-intercept form, but with f(x) instead of y.)
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Problem from today’s homework:
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Find the equation of the horizontal line through (1, 4).
• Recall that horizontal lines have an equation of the form y = c.
• So using the y-coordinate in the given point, y = 4.
Example
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Example
Find the equation of the vertical line through (-1,3).
• Recall that vertical lines have an equation of the form x = c.
• So using the x-coordinate in the given point, x = -1.
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Problem from today’s homework:
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Word problem similar to final problems in today’s homework:
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• Nonvertical parallel lines have identical slopes.
• Nonvertical perpendicular lines have slopes that are negative reciprocals of each other.
Remember: If you rewrite linear equations into slope-intercept form, you can easily determine slope to compare lines.
Parallel and perpendicular lines:
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• First, we need to find the slope of the given line.
3y = -x + 6 (subtract x from both sides)
Example:
3
1 y = x + 2 (divide both sides by 3)
• Since parallel lines have the same slope, we use the slope of for our new equation, together with the given point.
3
1
Find an equation of a line that contains the point (-2,4) and is parallel to the line x + 3y = 6. Write the equation in standard form.
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))2((3
14 xy
3( 4) 1( 2)y x (multiply by 3 to clear fractions)
2123 xy (use distributive property)
2123 yx (add x to both sides) (Why? Because they want it in STANDARD form)
103 yx (add 12 to both sides)
Example (cont.)
What would this look like in slope-intercept form?
In function notation?
1 10
3 3y x
1 10( )
3 3f x x
)( 11 xxmyy
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Find an equation of a line that contains the point (3,-5) and is perpendicular to the line 3x + 2y = 7. Write the equation in slope-intercept form.
• First, we need to find the slope of the given line.
2y = -3x + 7 (subtract 3x from both sides)
Example
(divide both sides by 2)3 7
2 2y x
• Since perpendicular lines have slopes that are negative reciprocals of each other, we use the slope of for our new equation, together with the given point (3,-5).
3
2
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)3(3
2)5( xy
Example (cont.)
)3(2)5(3 xy
(multiply by 3 to clear fractions)
62153 xy
(use distributive property)
13 15 2 65 15y x
(subtract 15 from both sides)
3 2 21y x
What would this look like in function notation?
27
3y x
2( ) 7
3f x x
(divide both sides by 3)
)( 11 xxmyy
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Problem from today’s homework:
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Find the equation of the line parallel to y = -4, passing through the point (0,-3).
• The line y = -4 is a horizontal line (slope = 0).• If the new line is parallel to this horizontal line
y = -4, then it must also be a horizontal line.• So we use the y-coordinate of our point to find
that the equation of the line is y = -3.
• NOTE: Sketching a quick graph of the line y = -4 and the point (0,-3) can help you visualize the situation and make sure you are solving the problem correctly.
Example:
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Find the equation of the line perpendicular to x = 7, passing through the point (-5,0).
• The line x = 7 is a vertical line.• If the new line is perpendicular to the vertical line
x = 7, then it must be a horizontal line.• So we use the y-coordinate of our point to find that the
equation of the line is y = 0.• Again: Sketching a quick graph of the line x = 7 and
the point (-5,0) can help you visualize the situation and make sure you are solving the problem correctly.
Example
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Problem from today’s homework:
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Problem from today’s homework:
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Reminder:
The homework assignment on Section 3.5 is due
at the start of next class period.
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You may now OPEN
your LAPTOPSand begin working on the
homework assignment.