Analytic Geometry Period 4 6205

7/28/2019 Analytic Geometry Period 4 6205

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http://www.leroy.k12.ny.us/facultywebs/robinson/images/math.gif


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Table of Contents 2 Variable Systems 3 Variable Systems Linear InequalitiesRational Inequalities Absolute Inequalities Circle EquationsDistance Equations


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Solving Linear Inequalities

Think about how you've done inequalities on thenumber line. For instance, they'd ask you to graph

something like x> 2. How did you do it? You woulddraw your number line, find the "equals" part (in thiscase, x= 2), mark this point with the appropriatenotation (an open dot or a parenthesis, indicating that thepoint x= 2 wasn't included in the solution), and then you'dshade everything to the right, because "greater than" meant"everything off to the right". The steps for linear inequalitiesare very much the same.


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Graph the solution to y < 2x + 3.

Just as for number-line inequalities, the first step is to find the "equals

part. In this case, the "equals" part is the line y= 2x+ 3. There are a coupleways you can graph this: you can use a T-chart, or you can graph from the y

intercept and the slope. Either way, you get a line that looks like this:


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Now we're at the point where your book gets complicated,with talk of "test points" and such. When you did those one-variable inequalities (like x< 3), did you bother with "testpoints", or did you just shade one side or the other? Ignore

the "test point" stuff, and look at the original inequality: y< 2x+ 3. You've already graphed the "or equal to" part (it's just the

line); now you're ready to do the "y less than" part. In otherwords, this is where you need to shade one side of the lineor the other. Now think about it: If you need yLESS THANthe line, do you want ABOVE the line, or BELOW? Naturally,you want below the line. So shade it in:


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Graph the solution to 2x 3y < 6.

First, solve for y:

2x 3y< 63y< 2x+ 6

y> ( 2/3 )x 2

[Note the flipped inequality sign in the last line. Don't forget to flip the

inequality if you multiply or divide through by a negative!] Now you need tofind the "equals" part, which is the line y= ( 2/3 )x 2. It looks like this:


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But this is what is called a "strict" inequality. That is,

it isn't an "or equals to" inequality; it's only "ygreaterthan". When you had strict inequalities on the numberline (such as x< 3), you'd denote this by using aparenthesis (instead of a square bracket) or an open[unfilled] dot (instead of a closed [filled] dot). In thecase of these linear inequalities, the notation for a

strict inequality is a dashed line. So the border of thesolution region actually looks like this:


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By using a dashed line, you still know where

the border is, but you also know that itisn't included in the solution. Since this isa "ygreater than" inequality, you want toshade abovethe line, so the solution looks

like this:


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Rational

InequalitiesEdward Duong
http://www.leroy.k12.ny.us/facultywebs/robinson/images/math.gif


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Rational Inequalitiesx-1 . < 0

(x-2)(x+)

1.Factor numerator and denominator

2.Determine the crucial numbers

(crucial number are the restricted values and the values that will

make the numerator 0)

The crucial numbers here are 2, -3 , and 1

3. Place crucial numbers on a number line4. Test Each Interval

(- , -3)(1,2)


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Examples of Rational

InequalitiesE.Ix > 0 here it shows that x over x-5 is greater

X-5 than 0.

Now find out the crucial numbers which are the numbers that make

the numerator and denominator equal to 0. The crucial numbers

here are 0 and 5

Now we can put the crucial numbers on the number line.


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Example of Rational

InequalitiesE.I

x > 0 here it shows that x over x-5 is greaterX-5 than 0.

Now find out the crucial numbers which are the numbers

that make the numerator and denominator equal to 0. The

crucial numbers here are 0 and 5 Now we can put the crucial numbers on the number line.

Continued on the next page


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Example of Rational

Inequalities The 0 and 5 on the number line represents the

crucial numbers

Now test each of the intervals and you get your

answer

(- ,0),(5, )


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Absolute Inequalities


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Absolute-Value InequalitiesAbsolute value: "|x| is the distance ofxfrom zero." |2 | = | 2 | = 2

Example:

-Absolute-Value Inequalities : |x| < 3 :

-The number 1 will work, as will1; the number 2 will work, as will2. But 4 willnot work, and neither will4, because they are too far away. Even 3 and3 won'twork (though they're right on the edge). But 2.99 will work, as will2.99. In otherwords, all the points between3 and 3, but not actually including3 or 3, willwork in this inequality. Then the solution looks like this:

-Translating this into symbols, you find that the solution is3


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Absolute-Value InequalitiesSolve | 2x + 3 | < 6

a. |2x+3|


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Absolute-Value Inequalitiesx < 19 orx > 24

a.Take a look at the endpoints. Nineteen and 24 are five units apart. Half of fiveis 2.5. So you want to adjust the inequality so it relates to2.5 and 2.5, instead of19 and 24. Since 19

(2.5) =21.5 and 24 2.5 = 21.5, you'll need to subtract 21.5 all around:b. x< 19 or x> 24

x 21.5 < 19 21.5 or x 21.5 > 24 21.5x 21.5 2.5

c. Since this is the "greater than" format, the absolute-value inequality will be ofthe form "absolute value of something is greater than or equal to 2.5". You can

convert this nicely to | x 21.5 | > 2.5


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CirclesKevin Dao


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1 2 3 4-4 -3 -2 -1

4

3

21

-1

-2

-3

-4

y

x

Formula for a circle is:

(x - h) + (y - k) = ror

x + y = r

Center of circle = (h , k)

Radius = r

Equation for the graph on the leftis: (x-1) + (y-1) = 1Center = (1,1)

Radius = 1

Ex. x + y - 2x + 4y + 3 = 0

(x - 2x) + (y + 4y) = -3

(x - 2x + 1) + (y + 4y + 4) = -3 + + 1 + 4

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

Center = (1 , -2)

Radius = 2


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DistanceKevin Dao


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Steps to finding distance

1. You must first find the horizontaldistance

-by using the x intercepts and findingthe differences between each line,

which in this case would be 2- (-1)

= 3 units

2. You must find the vertical distance-by using the y intercepts and findingthe differences between each line,

which in this case would be 2- (-

4) = 6 units

Continued on next slide.

x

y

l= 2y= 4x + 4

l= y=2x - 4

l

l


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Steps to finding distance

Continued3. You must now find the shortest distance using the distance formulaA x + B y + C

A + B- You must pick 1 point on each line and use thedistance formula on eachof the points.

P (x , y)

(1 , 4) 2(1) + (-1)(4)-4y= 2x 2 2+(-1)= 2x y 4 =6 . 5 65

A B C 5 5 5

P(1 , -2) 2(1) + (-1)(-2)+2y= 2x + 2 2+(-1)= 2x y + 2 = 6 . 5 65

5 5 5

=

=

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Analytic Geometry Period 4 6205