Intersections of Graphs in Visual Studio .NET

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Intersections of Graphs
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The intersection of two graphs consists of the points that the graphs have in common. These points can be found by solving simultaneously the equations that determine the graphs. EXAMPLES
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(a) To nd the intersection of the lines L and M determined by L : 4x 3y = 15 multiply the rst equation by 2 and the second equation by 3, 8x 6y = 30 9x + 6y = 21 Now y has the coef cients 6 and 6 in the two equations, and we add the equations to eliminate y, 17x = 51 or x = 51 =3 17 M : 3x + 2y = 7
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From the equation for M , when x = 3, 3(3) + 2y = 7. Hence, 9 + 2y = 7 or 2y = 2 or y = 1
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Thus, the only point of intersection is (3, 1) (see Fig. 5-1).
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Fig. 5-1
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CHAP. 5]
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INTERSECTIONS OF GRAPHS
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(b) Let us nd the intersection of the line L : y = 2x + 1 and the circle (x 1)2 + (y + 1)2 = 16. We must solve the system y = 2x + 1 (x 1)2 + (y + 1)2 = 16 By (1), substitute 2x + 1 for y in (2), (x 1)2 + (2x + 2)2 = 16 (x 2 2x + 1) + (4x 2 + 8x + 4) = 16 5x 2 + 6x + 5 = 16 5x 2 + 6x 11 = 0 (5x + 11)(x 1) = 0 Hence, either 5x + 11 = 0 or x 1 = 0; that is, either x = 11/5 = 2.2 or x = 1. By (1), when x = 1, y = 3; and when x = 2.2, y = 3.4. Thus, there are two intersection points, (1, 3) and ( 2.2, 3.4), as indicated in Fig. 5-2. (c) To nd the intersection of the line y = x + 2 and the parabola y = x 2 , we solve the system y =x+2 y = x2 By (4), substitute x 2 for y in (3), x2 = x + 2 x2 x 2 = 0 (x 2)(x + 1) = 0 Hence, either x 2 = 0 or x + 1 = 0; that is, either x = 2 or x = 1. By (3) or (4), when x = 2, y = 4; and when x = 1, y = 1. Thus, the intersection points are (2, 4) and ( 1, 1) (see Fig. 5-3). (3) (4) (1) (2)
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Fig. 5-2
Fig. 5-3
Solved Problems
5.1 Find the intersection of the lines L: 3x 3y = 1 M: 4x + 2y = 3
INTERSECTIONS OF GRAPHS
[CHAP. 5
We must solve the system 3x 3y = 1 4x + 2y = 3 In order to eliminate y, multiply the rst equation by 2 and the second by 3, 6x 6y = 2 12x + 6y = 9 Add the two equations, 18x = 11
11 Substitute 18 for x in the rst equation,
11 18
11 18
3y = 1 11 1 = 3y 6 5 = 3y 6 5 =y 18
11 5 So, the point of intersection is 18 , 18 .
5.2 Find the intersection of the line L : y = x + 3 and the ellipse
y2 x2 + = 1. 9 4
To solve the system of two equations, substitute x + 3 (as given by the equation of L ) for y in the equation of the ellipse, x2 (x + 3)2 + =1 9 4 Multiply by 36 to clear the denominators, 4x 2 + 9(x + 3)2 = 36 4x 2 + 9(x 2 + 6x + 9) = 36 4x 2 + 9x 2 + 54x + 81 = 36 13x 2 + 54x + 45 = 0
algebra The solutions of the quadratic equation ax 2 + bx + c = 0 are given by the quadratic formula x= By the quadratic formula, x= 54 54 (54)2 4(13)(45) = 2(13) 54 576 54 24 2916 2340 = = 26 26 26 b b2 4ac 2a
CHAP. 5]
INTERSECTIONS OF GRAPHS
Hence, either x= or x= 30 15 54 + 24 = = 26 26 13 78 54 24 = = 3 26 26 and and y =x+3= 15 39 24 + = 13 13 13
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