Unit 5, Lesson 3

Do Now Which quadrant are each of the following points located- 1.(-2, 3) Quadrant II 2.(9, 7) Quadrant I 3.(-4, -10.6) Quadrant III

HW & Objectives Be able to graph: lines, parabolas, absolute value Be able to find x and y intercepts Be able to determine the equation of a circle Vocabulary: parabola, x-intercepts, y-intercepts, diameter, radius, (x – h) 2 + (y – k) 2 = r 2 HW: Read p. 147 – 151 Do: p. 154: 2, 3, 5, 9, 11, 13, 17, 23, 37, 41, 43, 45, 49, 51, 53

Review Find the Fundamental Principle of Analytic Geometry on p Answer Question 1 on p. 154 replace x with 2 and y with 3; No 2. Explain what it means in your own words.

New: Finding X & Y Intercepts Key Fact: All x-intercepts have 0 as the y- coordinate; all y-intercepts have 0 as the x- coordinate For example (0,2); (0, -7); (0, 12) & (0, -7.8) are all y-intercepts since the x-coordinate is 0 Give four examples of coordinates for x- intercepts Possible answers: (4,0); (-1,0);(.05, 0);(-3,0)

New: Finding X & Y Intercepts 1.To find the x & y intercepts using the graph, simply locate where the graph crosses the x & y axes. 2.To find the x-intercept using an equation: a) replace y with 0 b) solve for x c) the coordinates for the x-intercept will be (x,0)

New: Finding X & Y Intercepts 3.Example: Find the x-intercepts of the graph of the equation y = x 2 – 6 a) replace y with 0: 0 = x 2 – 6 b) solve for x: 0 = x 2 – 6  6 = x 2  +√6 = x c) the x-intercept has coordinates (+√6, 0)

New: Finding X & Y Intercepts Find the y-intercept for the graph of the equation 2x + 3y = 14 a)Replace x with 0. Why? 2(0) + 3y =14; all y-intercepts have 0 as an x coordinate b)Solve for y: 0 + 3y = 14  3y = 14  y = 14/3 c)Coordinates for the y-intercept are (0, 14/3)

New: Equations for Circles 1.Given the equation of a line such as y = 2x – 8, you can graph the line. You’d find two points and connect them. 2. Also, since specific equations yield specific graphs, you can start with the graph and derive the equation. 3. For example, the standard form (equation) for a circle is: (x – h) 2 + (y – k) 2 = r 2 in which the center of the circle is (h,k) and the radius is r. Given the graph or information about the graph, you can write the equation. 4.The equation for a circle with the origin as the center is x 2 + y 2 = r 2 since (x+0) 2 + (y + 0) 2 = r 2 is the same as x 2 + y 2 = r 2

New: Equations for Circles 5.Ex. What is the equation of a circle with a center of (6,-5) and a radius of 8? a)Replace h and k in the standard form with 6 & 5: (x – 6) 2 + (y – - 5) 2 = r 2  (x – 6) 2 + (y + 5) 2 = r 2 ; Why is 5 positive? b) Replace r with 8: (x – 6) 2 + (y + 5) 2 = 8 2  (x – 6) 2 + (y + 5) 2 = 64

New: Equations for Circles Ex. Find the center and the radius of each of the following: 1. x 2 + (y – 7) 2 = 16 a)Find the center by putting into standard form: (x - 0) 2 + (y – 7) 2 = r 2 the center is (0,7) b)Find the radius by taking the sqrt of 16 since r 2 = 16 the radius is 4

New: Equations for Circles 2. Find the center and radius of the circle whose graph has the equation of (x + 2) 2 + (y -3) 2 = 18 a) Put in standard form: (x - - 2) 2 + (y – 3) 2 = r 2 the center is ( - 2,3) b) r 2 = 18; find sqrt of 18 to solve for r r = √ 18

Summary A.One way to graph an equation in two variables is to find some solutions to the equation and plot the points B.To find the x or y intercepts for a graph, plug in 0 for x to find the y-int and 0 for y to find the x-int C.(x – h) 2 + (y – k) 2 = r 2 is the equation for a circle; (h,k) is the center and r is the radius.