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Stitz-Zeager_College_Algebra_e-book

# To that end we make good use of the techniques

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Unformatted text preview: ns and now turn to verifying −2 our solution graphically. We begin by solving x2 +2xy − 16 = 0 for y to obtain y = 162xx . This function is easily graphed using the techniques of Section 4.2. Solving the second equation, y 2 + 2xy − 16 = √ for y , however, is more complicated. We use the quadratic formula to 0, obtain y = −x ± x2 + 16 which would require the use of Calculus or a calculator to graph. Believe it or not, we don’t need either because the equation y 2 + 2xy − 16 = 0 can be obtained from the equation x2 + 2xy − 16 = 0 by interchanging y and x. Thinking back to Section 5.2, this means we can obtain the graph of y 2 + 2xy − 16 = 0 by reﬂecting the graph of x2 + 2xy − 16 = 0 across the line y = x. Doing so conﬁrms that the two graphs intersect twice: once in Quadrant I, and once in Quadrant III as required. y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 The graphs of x2 + 2xy − 16 = 0 and y 2 + 2xy − 16 = 0 2. Unlike the previous problem, there...
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