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

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

1 Digital Lesson on 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 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 Plot the points in the coordinate plane –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 Connect the points with a straight line –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 –2 Example: Sketch Graph (Quadratic Function)

8 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8 Example: Graph

9 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9 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 – Example: Sketch Graph (Absolute Value Function)

10 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10 The points at which the graph intersects the the x- or y-axis are called intercepts. A point at which the graph of an equation meets the y-axis is called a y-intercept. It is possible for a graph to have no intercepts, one intercept, or several 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 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

12 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 12 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) – 5 = –5 So, the y-intercept is (0, –5). Example: Find Intercepts

13 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 13 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

14 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 14 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

15 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 15 Graphical Tests for Symmetry 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. As an illustration of this we graph y = x 2

16 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 16 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. As an illustration of this we graph y 2 = x.

17 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 17 Graphical Tests for Symmetry A graph is symmetric with respect to the origin if, whenever (x, y) is on the graph, (-x, -y) is also on the graph. As an illustration of this we graph y = x 3

18 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 18 Algebraic Tests for Symmetry The graph of an equation is symmetric with respect to the y-axis if replacing x with –x yields an equivalent equation. The graph of an equation is symmetric with respect to the x-axis if replacing y with –y yields an equivalent equation. The graph of an equation is symmetric with respect to the origin if replacing x with –x and replacing y with –y yields an equivalent equation. The algebraic tests for symmetry are as follows:

19 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 19 Algebraic Tests for Symmetry Example. The graph of y = x3 – x is symmetric with respect to the origin because:

20 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 20 Circles A circle with center at (h, k) and radius r consists of all points (x, y) whose distance from (h, k) is r. From the Distance Formula, we have the standard equation of a circle as:

21 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 21 Circles Example. Find the standard form of the equation of the circle with center at (2, -5) and radius 4. (x-2) 2 +(y-(-5)) 2 =4 2 or (x-2) 2 +(y+5) 2 =16


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