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# Chapter13 - Chapter 13 Joy Zhou November 4 2009 Chapter 13...

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Chapter 13 Joy Zhou November 4, 2009 Chapter 13 13.1, 13.2, 13.3, 13.4 13.1 Graph of the four lines are sketched in the back of the book. (a) The unit circle has equation x 2 + y 2 = 1 . The intersection(s) of the unit circle with the line y = - x + 2 is the solution to the equation set braceleftBigg x 2 + y 2 = 1 , y = - x + 2 (1) Plug y = - x + 2 into x 2 + y 2 = 1 . That gives us the equation x 2 + ( - x + 2) 2 = 1 . (2) 2 x 2 - 2 2 x + 1 . The discriminator of this quadratic equation is Δ = (2 2) 2 - 4 · 2 · 1 = 0. So it has a unique solution. Same idea for the other three lines. 1

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(b) From equation (2), we know x = 2 / 2 for the intersection of the unit circle and the line y = - x + 2. Plug into either the line equation to get y = - 2 / 2. So this intersection point is ( 2 / 2 , - 2 / 2). Same idea for the other three lines. (c) Move every line down by 1 unit and left by 2 units. Using the shifting techniques in section 13.3, the new line equations should be, accordingly, y - ( - 1) = - ( x + 2) + 2 (3) y - ( - 1) = - ( x + 2) - 2 (4) y - ( - 1) = ( x + 2) + 2 (5) y - ( - 1) = ( x + 2) - 2 (6) 13.2 Graphes sketched in the back of the book. (a) shift to the left by 1 unit, and vertically dilate by 2. (b) horizontally dilate by 1/2, then shift to the right by 1 unit, and then
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