SECTION 1.3 Intercepts; Symmetry; Graphing Key Equations 25 To graph the equation = x y2 using most graphing utilities, write the equation in the form { } = y x expression in . We proceed to solve for y. = = = ± x y y x y x 2 2 Square Root Method To graph = x y ,2 graph both = Y x 1 and = − Y x 2 on the same screen. Figure 39 shows the result using a TI-84 Plus CE.Table 8 shows various values of y for a given value of x when = Y x 1 and = − Y x. 2 Notice that when < x 0 we get an error. Can you explain why? Figure 38 x y2 = x y (9, 3) 6 (4, 2) (0, 0) (1, 21) 10 22 (9, 23) 5 (1, 1) (4, 22) Figure 39 22 26 6 10 Y2 5 2!x Y1 5 !x Table 8 (b) If we restrict y so that ≥ y 0, the equation = ≥ x y y , 0, 2 may be written as = y x. The portion of the graph of = x y2 in quadrant I plus the origin is the graph of = y x. See Figure 40. Figure 40 y x = x y (9, 3) 6 (4, 2) (0, 0) 10 5 (1, 1) 22 6 26 22 10 Y1 5 !x x y x 1 = x y, ( ) 1 10 10 1 10 , 10 ( ) 1 3 3 1 3 , 3 ( ) 1 2 2 1 2 , 2 ( ) 1 1 1, 1 ( ) 2 1 2 2, 1 2 ( ) 3 1 3 ( ) 3, 1 3 10 1 10 ( ) 10, 1 10 Table 9 Graphing the Equation = y x 1 Graph the equation = y x 1 . First, find any intercepts and check for symmetry. Solution EXAMPLE 9 Check for intercepts first. If we let = x 0, we obtain 0 in the denominator, which makes y undefined. We conclude that there is no y-intercept. If we let = y 0, we get the equation = x 1 0, which has no solution. We conclude that there is no x-intercept. The graph of = y x 1 does not cross or touch the coordinate axes. Next check for symmetry: x-Axis: Replacing y by −y yields − = y x 1 , which is not equivalent to = y x 1 . y-Axis: Replacing x by −x yields y x x 1 1 , = − = − which is not equivalent to = y x 1 . Origin: Replacing x by −x and y by −y yields y x 1 , − = − which is equivalent to = y x 1 . The graph is symmetric with respect to the origin. Now set up Table 9, listing several points on the graph. Because of the symmetry with respect to the origin, we use only positive values of x. From Table 9 we infer that (continued)
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