Digital Lesson on Graphs of Equations The graph

Digital Lesson on Graphs of Equations

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 2 y – x = 7 because the equation is satisfied when – 1 is substituted for x and 3 is substituted for y. That is, 2 y – x = 7 2(3) – (– 1) = 7 7=7 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Original Equation Substitute for x and y. Equation is satisfied. 2

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. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3

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

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

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

Example: Sketch the graph of y = (x – 1)2. x – 2 – 1 0 1 2 3 4 y 9 4 1 0 1 4 9 (x, y) (– 2, 9) (– 1, 4) (0, 1) (1, 0) (2, 1) (3, 4) (4, 9) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. y 8 6 2 x – 2 2 4 7

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

Example: Sketch the graph of y = | x | + 1. x – 2 – 1 0 1 2 y 3 2 1 2 3 (x, y) (– 2, 3) (– 1, 2) (0, 1) (1, 2) (2, 3) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. y 4 2 x – 2 2 9

The points at which the graph intersects the xor 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. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10

Procedure for finding the x- and y- intercepts of the graph of an equation algebraically: 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. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 11

Example: Find the x- and y-intercepts of the graph of y = x 2 + 4 x – 5. To find the x-intercepts, let y = 0 and solve for x. 0 = x 2 + 4 x – 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 = 02 + 4(0) – 5 = – 5 So, the y-intercept is (0, – 5). Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 12

Procedure for finding the x- and y-intercepts of the graph of an equation graphically: To find the x-intercepts of the graph of an equation, locate the points at which the graph intersects the x-axis. To find the y-intercepts of the graph of an equation, locate the points at which the graph intersects the y-axis. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 13

Example: Find the x- and y-intercepts of the graph of x = | y | – 2 shown below. y The graph intersects the x-axis at (– 2, 0). The graph intersects the y-axis at (0, 2) and at (0, – 2). 2 x – 3 1 2 3 The x-intercept is (– 2, 0). The y-intercepts are (0, 2) and (0, – 2). Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 14

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 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 15

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. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 16

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 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 17

Algebraic Tests for Symmetry The algebraic tests for symmetry are as follows: • 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. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 18

Algebraic Tests for Symmetry Example. The graph of y = x 3 – x is symmetric with respect to the origin because: Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 19

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: Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 20

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=42 or (x-2)2+(y+5)2=16 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 21