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

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 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.


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.


Example: Sketch the graph of y = –2x + 3.

1. Find several solution points of the equation.

x y = –2x + 3 (x, y)–2 y = –2(–2) + 3 = 7 (–2, 7) –1 y = –2(–1) + 3 = 5 (–1, 5) 0 y = –2(0) + 3 = 3 (0, 3)1 y = –2(1) + 3 = 1 (1, 1)2 y = –2(2) + 3 = –1 (2, –1)



2. Plot the points in the coordinate plane.

4 8

4

8

4

–4

x

yx y (x, y)

–2 7 (–2, 7) –1 5 (–1, 5) 0 3 (0, 3)1 1 (1, 1)2 –1 (2, –1)



3. Connect the points with a straight line.

4 8

4

8

4

–4

x

y


Example: Sketch the graph of y = (x – 1)2.

x y (x, y)–2 9 (–2, 9) –1 4 (–1, 4)0 1 (0, 1)1 0 (1, 0)2 1 (2, 1)3 4 (3, 4)4 9 (4, 9)

y

x2 4

2

6

8

–2


Example: Sketch the graph of y = | x | + 1.

x y (x, y)–2 3 (–2, 3)–1 2 (–1, 2) 0 1 (0, 1)1 2 (1, 2)2 3 (2, 3)

y

x–2 2

2

4


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.


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.


Intercepts of a Graph


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:


Example: Find the x- and y-intercepts of the graph of y = x2 + 4x – 5.To find the x-intercepts, let y = 0 and solve for x.

0 = x2 + 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 = 02 + 4(0) – 5 = –5

So, the y-intercept is (0, –5).


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.


Example: Find the x- and y-intercepts of the graph of x = | y | – 2 shown below.

y

x1

2

–3 2 3

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).


Symmetry • x-Axis symmetry

• y-Axis symmetry

• Origin symmetry


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.


Example

The graph of is symmetric with respect to the y-axis because …

X – 3 – 2 – 1 0 1 2 3

Y = x2 – 2 7 2 – 1 – 2 – 1 2 7

(x,y) (– 3,7) (– 2,2) (– 1,– 1) (0,– 2) (1,– 1) (2,2) (3,7)

2 2y x


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.


Example

• Use symmetry to sketch the graph of

Y X = y2 + 1 (x,y)

0 1 (1,0)

1 2 (2,1)

2 5 (5,2)

2 1x y


Example

• Sketch the graph of y = x – 1.

X –2 –1 0 1 2 3 4

Y = |x –1| 3 2 1 0 1 2 3

(x,y) (–2,3) (–1,2) (0,1) (1,0) (2,1) (3,2) (4,3)


Circles A point (x, y) is on the circle if and only if its distance f

rom the center (h, k) is r. By the Distance Formula,

rkyhx 22


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

2 2 2( ) ( )x h y k r

2 2 2x y r


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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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