Do Now!!! Find the values of x that satisfy and explain how you found your solution. Solution: First, you must factor the numerator and denominator if possible. Next, you must determine the endpoints of the intervals of your solution, and test values in that area. A sign graph can help you. x + 3 x – 3 x + 2 x – 2 fcn Since the function is less than/equal to zero, our solutions are where the negatives are in the last “fcn” line. Note: x≠ -2,2 since these values would make the denominator zero. The solution is: [-3, -2) U (2, 3]

1.3 The Coordinate Plane x-axis y-axis Quadrant I Quadrant 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 from zero and b represents the vertical distance from zero.

Ex 1: Sketch the points in the xy-plane that satisfy 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 original inequality 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 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:

Ex 3: Sketch the points in the xy-plane that satisfy the inequalities: Note: Don’t forget how to solve absolute value problems. breaks up into 4 different inequalities, so you will graph four separate lines just from this part of the problem. We get… This gives us our equations for the four lines… Do not change direction of symbols and list smallest value to largest value from left to right.

Graph these lines, but don’t forget to determine if they are solid or dashed and shade accordingly to the two inequalities that you get from the absolute value part of this problem. Now, we have to finish the second part of the problem… Graph the lines y = -1 and y = 3, both lines 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. Ex: Find the distance between the points (1, 2) and (-2, 6)

Midpoint Formula:

Circles in the Plane The distance formula for points in a plane can also be used for finding 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) and radius r precisely when (x – h) 2 + (y – k) 2 = r 2 A circle whose center is at the origin and whose radius is 1 is called a “unit circle.”

Ex 4:Determine an equation for the circle of radius 3 centered at (-1,2). (x – h) 2 + (y – k) 2 = r 2 The center is (h,k) (x + 1) 2 + (y – 2) 2 = 9 The negative sign in the formula and the negative sign on the x-coordinate make it positive. r 2 = 9 Note: This is the standard form of a circle. (x + 1) 2 + (y – 2) 2 = 9 If we expand the binomials and combine like terms we get the equation of the same circle only in a different form.

Ex 5:Sketch the circle with equation When the equation of a circle is in this form, we cannot immediately see the center nor the radius. We must use a technique called Completing the Square to 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. 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. b = - 4 in the second quantity.

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. 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.

Center: (h,k) = (-1, 2) radius: x y