Stitz-Zeager_College_Algebra_e-book

# To get a better idea of the shape of the graph we

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Unformatted text preview: he graph of x2 + y 3 = 1. Solution. To check, we substitute x = 2 and y = −1 into the equation and see if the equation is satisﬁed ? (2)2 + (−1)3 = 1 3=1 Hence, (2, −1) is not on the graph of x2 + y 3 = 1. We could spend hours randomly guessing and checking to see if points are on the graph of the equation. A more systematic approach is outlined in the following example. Example 1.3.2. Graph x2 + y 3 = 1. Solution. To eﬃciently generate points on the graph of this equation, we ﬁrst solve for y x2 + y 3 y3 3 y3 y = = = = 1 1 − x2 √ 3 1 − x2 √ 3 1 − x2 We now substitute a value in for x, determine the corresponding value y , and plot the resulting point, (x, y ). For example, for x = −3, we substitute y= 3 1 − x2 = 3 1 − (−3)2 = √ 3 −8 = −2, so the point (−3, −2) is on the graph. Continuing in this manner, we generate a table of points which are on the graph of the equation. These points are then plotted in the plane as shown below. 1.3 Graphs of Equations 23 y x −3 −2 −1 0...
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