Do Now!!!

Find the values of x that satisfy and explain how youfound your solution.

049

2

2

xx

Solution: First, you must factor the numerator and denominatorif possible.

0)2)(2()3)(3(

xxxx

Next, you must determine the endpoints of the intervalsof your solution, and test values in that area. A sign graph can help you.

x + 3x – 3x + 2x – 2fcn

-3 -2 2 3

------0++++++++++++++++------------------------------0++-----------0+++++++++++++--------------------------0+++++++ 0---0++++++++ 0- -0++

Since the function is lessthan/equal to zero, oursolutions are where thenegatives are in the last“fcn” line. Note: x≠ -2,2 sincethese values would make thedenominator zero.

The solutionis: [-3, -2) U (2, 3]

1.3 The Coordinate Plane

x-axis

y-axis

Quadrant IQuadrant II

Quadrant III Quadrant IV

Points are locations in the coordinate plane that are represented by ordered pairs (a, b), where a represents the horizontal distance fromzero and b represents the vertical distance from zero.

Ex 1: Sketch the points in the xy-plane that satisfy 12 yandx

Solution: •Remove the inequality symbols and replace them with equal signs.

x = 2 and y = 1

•Graph those lines on the coordinate plane.

y

x

2

1

Note: if the originalinequality is not“equal to” it is a dashed line. If it is“equal to” it is a solid line.

•Now shade. Since there are two inequalities in this problem, both have to be satisfied at the same time.

•For x > 2, shade where the x’s would be greater than 2.- to the right.

•For 1y shade where y is less than one.

- below.

The area where the two shadings overlap are your solutions.

Do Now!!!

Graph the following in a coordinate plane: 12 xy

Ex 3: Sketch the points in the xy-plane that satisfy the inequalities:

3121 yandx

Note: Don’t forget how to solve absolute value problems.

21 x breaks up into 4 different inequalities, so you willgraph four separate lines just from this part of the problem.

We get… 1221 xorx

This gives us our equations for the four lines…

2

2

1

1

x

x

x

x

2

2

1

1

x

x

x

x

Do not change direction of symbols and list smallest value to largest value from left to right.

2

2

1

1

x

x

x

x Graph these lines, but don’t forget to determine if theyare solid or dashed and shade accordingly to the twoinequalities that you get from the absolute value partof this problem. 1221 xorx

Now, we have to finish the second partof the problem… 31 y

Graph the lines y = -1 and y = 3, bothlines will be dashed.

Shade above y = -1 and below y = 3.

Your solutions are where the shadings overlap.

Distance between two points in a plane:

The formula for finding the distance between two points in a coordinate plane is found by using a variation of the Pythagorean Theorem.

2

21

2

212211)()(),(),,( yyxxyxyxd

Ex: Find the distance between the points (1, 2) and (-2, 6)

22 )62())2(1()6,2(),2,1( d

5)6,2(),2,1( d

Midpoint Formula:

)(

21

),(21

2121yyxx

Circles in the Plane

The distance formula for points in a plane can also be used forfinding the equation of a circle.

Circle:

A circle is a set of all points whose distance from a given point, the center, is a fixed distance, the radius.

The point (x,y) lies on the circle with center (h,k) andradius r precisely when (x – h)2 + (y – k)2 = r2

A circle whose center is at the origin and whose radius is 1is called a “unit circle.”

Ex 4: Determine an equation for the circle of radius 3 centeredat (-1,2).

(x – h)2 + (y – k)2 = r2 The center is (h,k)

(x + 1)2 + (y – 2)2 = 9 The negative sign in the formula and the negative signon the x-coordinate make itpositive. r2 = 9

Note: This is the standard form of a circle.

(x + 1)2 + (y – 2)2 = 9 If we expand the binomials and combinelike terms we get the equation of the same circle only in a different form.

94412 22 yyxx 44222 yxyx

44222 yxyxEx 5: Sketch the circle with equation

When the equation of a circle is in this form, we cannotimmediately see the center nor the radius.

We must use a technique called Completing the Squareto determine the center and radius.

Step 1: Group like terms together with parentheses and put the constant(s) on the other side of the equation.

4)4()2( 22 yyxx

Step 2: To complete the square, you work with one quantity at a time. Take the value of “b”, divide it by 2 and square the result.

b = 2 in the first quantity.1

2

22

b = - 4 in the second quantity. 42

42

Step 3: Add those two new values in the open spaces.

Note: Don’t forget! What you do to one side of an equation you must do to the other.

9)2()1(

9)2)(2()1)(1(

9)44()12(

22

22

yx

yyxx

yyxx

Step 4: Factor your quadratics. If you completed the square correctly, you will get the same binomial when you factor. (not necessarily the same in each quadratic)

Now, the equation is in standard form and you can determine the center and radius so that you can graph.

9)44()12( 22 yyxx

9)2()1( 22 yx Center: (h,k) = (-1, 2)radius: 3

-1 2-4

2

5

-1

x

y