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Digital Lesson Graphs of Equations. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 The graph of an equation in two variables x and.

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Presentation on theme: "Digital Lesson Graphs of Equations. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 The graph of an equation in two variables x and."— Presentation transcript:

1 Digital Lesson Graphs of Equations

2 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 The graph of an equation in two variables x and y is the set of all points (x, y) whose coordinates satisfy the equation. For instance, the point (–1, 3) is on the graph of 2y – x = 7 because the equation is satisfied when –1 is substituted for x and 3 is substituted for y. That is, 2y – x = 7 Original Equation 2(3) – (–1) = 7 Substitute for x and y. 7 = 7 Equation is satisfied. Definition of Graph

3 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3 To sketch the graph of an equation, 1.Find several solution points of the equation by substituting various values for x and solving the equation for y. 2. Plot the points in the coordinate plane. 3.Connect the points using straight lines or smooth curves. Sketching Graphs

4 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 Example: Sketch the graph of y = –2x + 3. 1. Find several solution points of the equation. xy = –2x + 3(x, y) –2y = –2(–2) + 3 = 7(–2, 7) –1y = –2(–1) + 3 = 5(–1, 5) 0y = –2(0) + 3 = 3(0, 3) 1y = –2(1) + 3 = 1(1, 1) 2y = –2(2) + 3 = –1(2, –1) Example: Sketch Graph (Linear Function)

5 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5 Example: Sketch the graph of y = –2x + 3. 2. Plot the points in the coordinate plane. 48 4 8 4 –4 x y xy(x, y) –27(–2, 7) –15(–1, 5) 03(0, 3) 11(1, 1) 2–1(2, –1) Example continued

6 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 6 Example: Sketch the graph of y = –2x + 3. 3. Connect the points with a straight line. 48 4 8 4 –4 x y Example continued

7 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 7 Example: Sketch the graph of y = (x – 1) 2. xy(x, y) –29(–2, 9) –14(–1, 4) 01(0, 1) 10(1, 0) 21(2, 1) 34(3, 4) 49(4, 9) y x 24 2 6 8 –2 Example: Sketch Graph (Quadratic Function)

8 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8 Example: Sketch the graph of y = | x | + 1. xy(x, y) –23(–2, 3) –12(–1, 2) 01(0, 1) 12(1, 2) 23(2, 3) y x –22 2 4 Example: Sketch Graph (Absolute Value Function)

9 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9 The point-plotting technique demonstrated in above Example is easy to use, but it has some shortcomings. With too few solution points, you can misrepresent the graph of an equation. Definition of Intercepts

10 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10 The points at which the graph intersects the x-axis or y-axis are called intercepts. If (x, 0) satisfies an equation, then the point (x, 0) is called an x-intercept of the graph of the equation. If (0, y) satisfies an equation, then the point (0, y) is called a y-intercept of the graph of the equation. Definition of Intercepts

11 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 11 Intercepts of a Graph

12 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 12 To find the x-intercepts of the graph of an equation, substitute 0 for y in the equation and solve for x. To find the y-intercepts of the graph of an equation algebraically, substitute 0 for x in the equation and solve for y. Procedure for finding the x- and y- intercepts of the graph of an equation algebraically: Finding Intercepts Algebraically

13 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 13 Example: Find the x- and y-intercepts of the graph of y = x 2 + 4x – 5. To find the x-intercepts, let y = 0 and solve for x. 0 = x 2 + 4x – 5 Substitute 0 for y. 0 = (x – 1)(x + 5) Factor. x – 1 = 0 x + 5 = 0 Set each factor equal to 0. x = 1 x = –5 Solve for x. So, the x-intercepts are (1, 0) and (–5, 0). To find the y-intercept, let x = 0 and solve for y. y = 0 2 + 4(0) – 5 = –5 So, the y-intercept is (0, –5). Example: Find Intercepts

14 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 14 To find the x-intercepts of the graph of an equation, locate the points at which the graph intersects the x-axis. Procedure for finding the x- and y-intercepts of the graph of an equation graphically: To find the y-intercepts of the graph of an equation, locate the points at which the graph intersects the y-axis. Finding Intercepts Graphically

15 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 15 Example: Find the x- and y-intercepts of the graph of x = | y | – 2 shown below. y x 1 2 –323 The x-intercept is (–2, 0). The y-intercepts are (0, 2) and (0, –2). The graph intersects the x-axis at (–2, 0). The graph intersects the y-axis at (0, 2) and at (0, –2). Example: Find Intercepts

16 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 16 Symmetry x-Axis symmetry y-Axis symmetry Origin symmetry

17 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 17 Graphical Tests for Symmetry A graph is symmetric with respect to the x-axis if, whenever (x,y) is on the graph, (x,-y) is also on the graph. A graph is symmetric with respect to the y-axis if, whenever (x,y) is on the graph, (-x,y) is also on the graph. A graph is symmetric with respect to the origin if, whenever (x,y) is on the graph, (-x,-y) is also on the graph.

18 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 18 Example The graph of is symmetric with respect to the y-axis because … X– 3– 2– 10123 Y = x 2 – 272– 1– 2– 127 (x,y)(– 3,7)(– 2,2)(– 1,– 1)(0,– 2)(1,– 1)(2,2)(3,7)

19 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 19 Algebraic tests for symmetry The graph of an equation is symmetric with respect to the x-axis if replacing y with –y yields equivalent equation. The graph of an equation is symmetric with respect to the y-axis if replacing x with –x yields equivalent equation. The graph of an equation is symmetric with respect to the origin if replacing x with –x and y with –y yields equivalent equation.

20 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 20 Example Use symmetry to sketch the graph of YX = y 2 + 1 (x,y) 01(1,0) 12(2,1) 25(5,2)

21 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 21 Example Sketch the graph of y =  x – 1 . X–2–101234 Y = |x –1|3210123 (x,y)(–2,3)(–1,2)(0,1)(1,0)(2,1)(3,2)(4,3)

22 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 22 Circles A point (x, y) is on the circle if and only if its distance from the center (h, k) is r. By the Distance Formula,

23 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 23 Standard form of the equation of a circle The point (x,y) lies on the circle of radius r and center (h,k) if and only if From this result, you can see that the standard form of the equation of circle with its center at the origin, (x – h) = (0,0), is simply

24 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 24 Example 1. The point (3,4) lies on a circle whose center is at (-1,2). Write the standard form of the equation of this circle.


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