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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
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Digital Lesson
Graphs of Equations
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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.
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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.
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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)
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Example: Sketch the graph of y = –2x + 3.
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)
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Example: Sketch the graph of y = –2x + 3.
3. Connect the points with a straight line.
4 8
4
8
4
–4
x
y
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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
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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
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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.
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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.
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Intercepts of a Graph
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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:
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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).
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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.
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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).
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Symmetry • x-Axis symmetry
• y-Axis symmetry
• Origin symmetry
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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.
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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
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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.
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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
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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)
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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
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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
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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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