# Graphs of Equations in .NET framework Drawer QR Code in .NET framework Graphs of Equations

Graphs of Equations
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Consider the following equation involving the variables x and y: 2y 3x = 6 (i)
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Notice that the point (2, 6) satis es the equation; that is, when the x-coordinate 2 is substituted for x and the y-coordinate 6 is substituted for y, the left-hand side, 2y 3x, assumes the value of the right-hand side, 6. The graph of (i) consists of all points (a, b) that satisfy the equation when a is substituted for x and b is substituted for y. We tabulate some points that satisfy (i) in Fig. 3-1(a), and indicate these points in Fig. 3-1(b). It is apparent that these points all lie on a straight line. In fact, it will be shown later that the graph of (i) actually is a straight line.
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Fig. 3-1 In general, the graph of an equation involving x and y as its only variables consists of all points (x, y) satisfying the equation.
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GRAPHS OF EQUATIONS
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[CHAP. 3
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EXAMPLES
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(a) Some points on the graph of y = x 2 are computed in Fig. 3-2(a) and shown in Fig. 3-2(b). These points suggest that the graph looks like what would be obtained by lling in the dashed curve. This graph is of the type known as a parabola.
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Fig. 3-2
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(b) The graph of the equation xy = 1 is called a hyperbola. As shown in Fig. 3-3(b), the graph splits into two separate pieces. The points on the hyperbola get closer and closer to the axes as they move farther and farther from the origin.
Fig. 3-3
CHAP. 3]
GRAPHS OF EQUATIONS
(c) The graph of the equation y2 x2 + =1 9 4 is a closed curve, called an ellipse (see Fig. 3-4).
Fig. 3-4 Circles For a point P(x, y) to lie on the circle with center C(a, b) and radius r, the distance PC must be r (Fig. 3-5). Now by (2.1), PC =
(x a)2 + (y b)2
The standard equation, PC = r 2 , of the circle with center (a, b) and radius r is then (x a)2 + (y b)2 = r 2 For a circle centered at the origin, (3.1) becomes simply x 2 + y2 = r 2 (3.2) (3.1)
Fig. 3-5
GRAPHS OF EQUATIONS
[CHAP. 3
EXAMPLES
(a) The circle with center (1, 2) and radius 3 has the equation (x 1)2 + (y 2)2 = 9 (b) The circle with center ( 1, 4) and radius 6 has the equation (x + 1)2 + (y 4)2 = 36 (c) The graph of the equation (x 3)2 + (y 7)2 = 16 is a circle with center (3, 7) and radius 4. (d) The graph of the equation x 2 + (y + 2)2 = 1 is a circle with center (0, 2) and radius 1.
Sometimes the equation of a circle will appear in a disguised form. For example, the equation x 2 + y2 6x + 2y + 6 = 0 is equivalent to (x 3)2 + (y + 1)2 = 4
algebra Use the formulas (u + v)2 = u2 + 2uv + v 2 and (u v)2 = u2 2uv + v 2 to expand the left-hand side of (iii).
(ii)
(iii)
If an equation such as (ii) is given, there is a simple method for recovering the equivalent standard equation of the form (iii) and thus nding the center and the radius of the circle. This method depends on completing the squares; that is, replacing the quantities x 2 + Ax and y2 + By by the equal quantities x+ A 2
A2 4
B2 4
EXAMPLE Let us nd the graph of the equation
x 2 + y2 + 4x 2y + 1 = 0 Completing the squares, replace x 2 + 4x by (x + 2)2 4 and y2 2y by (y 1)2 1, (x + 2)2 4 + (y 1)2 1 + 1 = 0 (x + 2)2 + (y 1)2 = 4 This is the equation of a circle with center ( 2, 1) and radius 2.