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Stitz-Zeager_College_Algebra_e-book

# If the result is equivalent to the original equation

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Unformatted text preview: and y : • the x-intercepts always have the form (x, 0); to ﬁnd the x-intercepts of the graph, set y = 0 and solve for x. • y -intercepts always have the form (0, y ); to ﬁnd the y -intercepts of the graph, set x = 0 and solve for y . Another fact which you may have noticed about the graph in the previous example is that it seems to be symmetric about the y -axis. To actually prove this analytically, we assume (x, y ) is a generic point on the graph of the equation. That is, we assume x2 + y 3 = 1. As we learned in Section 1.1, the point symmetric to (x, y ) about the y -axis is (−x, y ). To show the graph is symmetric about the y -axis, we need to show that (−x, y ) is on the graph whenever (x, y ) is. In other words, we need to show (−x, y ) satisﬁes the equation x2 + y 3 = 1 whenever (x, y ) does. Substituting gives ? (−x)2 + (y )3 = 1 x2 + y 3 = 1 When we substituted (−x, y ) into the equation x2 + y 3 = 1, we obtained the original equation back when we simpliﬁed. Th...
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