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3.1 Find the graph of: (a) the equation x = 2; (b) the equation y = 3.
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(a) The points satisfying the equation x = 2 are of the form (2, y), where y can be any number. These points form a vertical line [Fig. 3-6(a)]. (b) The points satisfying y = 3 are of the form (x, 3), where x is any number. These points form a horizontal line [Fig. 3-6(b)].
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GRAPHS OF EQUATIONS
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Fig. 3-6 3.2 Find the graph of the equation x = y2 .
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Plotting several points suggests the curve shown in Fig. 3-7. This curve is a parabola, which may be obtained from the graph of y = x 2 (Fig. 3-2) by switching the x- and y-coordinates.
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Fig. 3-7 3.3 Identify the graphs of: (a) 3x 2 + 3y2 6x y + 1 = 0 (c) x 2 + y2 + 20x 4y + 120 = 0
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(a) First, divide both sides by 3, 1 1 x 2 + y2 2x y + = 0 3 3 Complete the squares, (x 1)2 + y (x 1)2 + y 1 1 2 1 =0 + 1 6 3 36 1 36 1 12 25 1 2 1 = + = =1+ 6 36 3 36 36 36 36
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x 2 + y2 8x + 16y + 80 = 0
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Hence, the graph is a circle with center (1, 1 ) and radius 5 . 6 6
GRAPHS OF EQUATIONS
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(b) Complete the squares, (x 4)2 + (y + 8)2 + 80 16 64 = 0 or (x 4)2 + (y + 8)2 = 0
Since (x 4)2 0 and (y + 8)2 0, we must have x 4 = 0 and y + 8 = 0. Hence, the graph consists of the single point (4, 8). (c) Complete the squares, (x + 10)2 + (y 2)2 + 120 100 4 = 0 or (x + 10)2 + (y 2)2 = 16
This equation has no solution, since the left-hand side is always nonnegative. Hence, the graph consists of no points at all, or, as we shall say, the graph is the null set.
3.4 Find the standard equation of the circle centered at C(1, 2) and passing through the point P(7, 4).
The radius of the circle is the distance CP = (7 1)2 + [4 ( 2)]2 = 36 + 36 = 72
Thus, the standard equation is (x 1)2 + (y + 2)2 = 72.
3.5 Find the graphs of:
(a) y = x 2 + 2; (b) y = x 2 2; (c) y = (x 2)2 ; (d) y = (x + 2)2 .
(a) The graph of y = x 2 + 2 is obtained from the graph of y = x 2 (Fig. 3-2) by raising each point two units in the vertical direction [see Fig. 3-8(a)]. (b) The graph of y = x 2 2 is obtained from the graph of y = x 2 by lowering each point two units [see Fig. 3-8(b)]. (c) The graph of y = (x 2)2 is obtained from the graph of y = x 2 by moving every point of the latter graph two units to the right [see Fig. 3-8(c)]. To see this, assume (a, b) is on y = (x 2)2 . Then b = (a 2)2 . Hence, the point (a 2, b) satis es y = x 2 and therefore is on the graph of y = x 2 . But (a, b) is obtained by moving (a 2, b) two units to the right. (d) The graph of y = (x+2)2 is obtained from the graph of y = x 2 by moving every point two units to the left [see Fig. 3-8(d)]. The reasoning is as in part (c). Parts (c) and (d) can be generalized as follows. If c is a positive number, the graph of the equation F(x c, y) = 0 is obtained from the graph of F(x, y) = 0 by moving each point of the latter graph c units to the right. The graph of F(x + c, y) = 0 is obtained from the graph of F(x, y) = 0 by moving each point of the latter graph c units to the left.
3.6 Find the graphs of:
(a) x = (y 2)2 ; (b) x = (y + 2)2 .
(a) The graph of x = (y 2)2 is obtained by raising the graph of x = y2 [Fig. 3-7(b)] by two units [see Fig. 3-9(a)]. The argument is analogous to that for Problem 3.5(c). (b) The graph of x = (y + 2)2 is obtained by lowering the graph of x = y2 two units [see Fig. 3-9(b)]. These two results can be generalized as follows. If c is a positive number, the graph of the equation F(x, y c) = 0 is obtained from the graph of F(x, y) = 0 by moving each point of the latter graph c units vertically upward. The graph of F(x, y + c) = 0 is obtained from the graph of F(x, y) = 0 by moving each point of the latter graph c units vertically downward.
3.7 Find the graphs of:
(a) y = (x 3)2 + 2; (b) y (x 2) = 1.
(a) By Problems 3.5 and 3.6, the graph is obtained by moving the parabola y = x 2 three units to the right and two units upward [see Fig. 3-10(a)]. (b) By Problem 3.5, the graph is obtained by moving the hyperbola xy = 1 (Fig. 3-3) two units to the right [see Fig. 3-10(b)].
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