Stitz-Zeager_College_Algebra_e-book

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online