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Unformatted text preview: f A (0) = 0 f A ( e 1 ) = e 1 f A ( e 2 ) = 2 e 1 + 3 e 2 f A ( e 1 + e 2 ) = 3 e 1 + 3 e 2 . Thus the image is the quadrilateral with vertices (0 , 0) (1 , 0) (2 , 3) (3 , 3) , and you can now easily graph it. (b) The area of the triangle S is 3, as you can work out using the fact that area = 1 2 base × height . Since the magniﬁcation factor is 3, the area of the image is 9. To graph the image, we work out the vertices of the image: There are: f A ( e 1 + e 2 ) = 3 e 1 + 3 e 2 f A ( e 1 + 4 e 2 ) = 9 e 1 + 12 e 2 f A (3 e 15 e 2 ) =7 e 115 e 2 . Thus the image is the triangle with vertices (3 , 3) (9 , 12) (7 ,15) , 21 /september/ 2005; 22:09 46...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.
 Fall '08
 Gladue
 Calculus

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