# 7 10 pts very carefully sketch the graph of the

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provides a factorization of quadratics. 7. (10 pts.) Very carefully sketch the graph of the equation y 2 - x 2 = 1 below. [ See tr-t4g7.pdf. ] y x

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TEST-04/MAC1114 Page 4 of 4 8. (10 pts.) Find all the complex cube roots of 3 1/2 + i. Leave your answer in polar form with the arguments given in degrees. If z = 3 1/2 + i, then we can write z in polar form as z = 2[cos(30°) + sin(30°)i]. The three cube roots are w 0 = 2 1/3 [cos(10°) + sin(10°)i], w 1 = 2 1/3 [cos(130°) + sin(130°)i], and w 2 = 2 1/3 [cos(250°) + sin(250°)i] 9. (5 pts.) Find the vertex, focus, and directrix of the parabola that has the equation given below. y 2 - 4y = x + 4. By performing the usual algegraic magic you can transform the equation above into the standard form equation (y - 2) 2 = 4(1/4)(x - (-8)). Using this, you can easily see that the vertex is (-8,2), the focus is (-8 + (1/4), 2) = (-31/4, 2), and the directrix is the line defined by x = -8 - (1/4) = -33/4. 10. (5 pts.) Find the center, foci, and vertices of the ellipse that has the equation given below. 4x 2 + y 2 + 4y = 0. Again, by playing the complete-the-square game carefully, you should obtain the standard form equation (x - 0) 2 + (1/4)(y - (-2)) 2 = 1. From this you should have c = (4 - 1) 1/2 . Clearly the center is (0,-2), the two vertices are (0,0) and (0,-4), and the two foci are (0,2 + 3 1/2 ) and (0,2 - 3 1/2 ). 11. (5 pts.) Find the center, foci, and vertices of the hyperbola that has the equation given below. y 2 - 4x 2 -16x - 2y - 19 = 0 Finally, by being very puntilious in your algebraic prestidigitation, you can obtain the standard form equation (1/4)(y - 1) 2 - (x - (-2)) 2 = 1. The center is (-2,1), the two vertices are (-2,3) and (-2,-1), and since c = (4 + 1) 1/2 , the two foci are (-2, 1 + 5 1/2 ) and (-2, 1 - 5 1/2 ).
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