Ch 9/10/11/12 Exam Review

Transcription

1 Ch 9/0// Exam Review The vector v has initial position P and terminal point Q. Write v in the form ai + bj; that is, find its position vector. ) P = (4, 6); Q = (-6, -) Find the vertex, focus, and directrix of the parabola. Graph the equation. ) (y + ) = 4(x - ) Solve the problem. ) If v = 7i + 8j, what is v? Find the quantity if v = i - 7j and w = i + j. ) v - w Find the unit vector having the same direction as v. 4) v = i + 4j Find the dot product v w. ) v = 6i - j, w = 8i + j Find the angle between v and w. Round your answer to one decimal place, if necessary. 6) v = -i + 7j, w = -6i - 4j Solve the problem. 7) Which of the following vectors is parallel to v = -0i - 8j? A) w = 4i + 4j B) w = -0i + j C) w = i - j D) w = 0i + 6j State whether the vectors are parallel, orthogonal, or neither. 8) v = i + 4j, w = 4i - j Find the center, foci, and vertices of the ellipse. ) (x + ) 6 + (y + ) 6 = ) 64x + y - x + 0 = 0 Find an equation for the ellipse described. 4) Center at (, ); focus at (, ); vertex at (, ) Find an equation for the ellipse described. Graph the equation. ) Foci at (-, -) and (, -); length of major axis is 0 Find the vertex, focus, and directrix of the parabola with the given equation. 9) (y + ) = 0(x + ) Find an equation for the parabola described. 0) Vertex at (8, 6); focus at (, 6) Find an equation for the hyperbola described. 6) Vertices (, -) and (- 9, -); asymptotes y + = ± 6 (x + )

2 7) Vertices at (±, 0); foci at (±4, 0) Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola. 8) (x - ) - (y + ) 9 = Graph the equations of the system. Then solve the system to find the points of intersection. 4) y = x y = - x Graph the hyperbola. 9) (x + ) 4 - (y + ) = Write the partial fraction decomposition of the rational expression. x - 8 0) (x - )(x - ) x + ) (x - ) (x + 4) ) x - x 4 - x - 7 Solve the system of equations using substitution. ) 6) y = x - 4x + 4 x + y = 8 y = x - 4 y = -6x Solve using elimination. 7) x + y = 9 x - y = - ) x + x + x + 7 (x + ) 8) x + y = 7 x - y = -6 9) x + xy - y = x + xy + y = Evaluate the factorial expression. 0)!! Write out the first five terms of the sequence. ) {sn} = {(4n - )}

3 ) {cn} = (-) n (n + )(n + ) Find the sum. 4) The sequence is defined recursively. Write the first four terms. ) a = ; an = an- - Write out the sum. 4) n (k + ) k = 44) 4 n = (n - ) Find the fifth term and the nth term of the geometric sequence whose initial term, a, and common ratio, r, are given. 4) a = 9; r = Express the sum using summation notation. ) Find the nth term {an} of the geometric sequence. When given, r is the common ratio. 46) -4, -, -6, -08, -4,... Find the sum of the sequence. 6) 7 6k k = 4 Find the sum. 47) k= () k 7) (k + 0) k = Determine whether the infinite geometric series converges or diverges. If it converges, find its sum. 48) The given pattern continues. Write down the nth term of the sequence {an} suggested by the pattern. 8) 0,, 6,, 0,... Find the nth term and the indicated term of the arithmetic sequence whose initial term, a, and common difference, d, are given. 9) a = 0; d = an =?; a =? Find the indicated term of the sequence. 40) The tenth term of the arithmetic sequence -, -,,... 49) Use the Principle of Mathematical Induction to show that the statement is true for all natural numbers n. 0) (n - ) = n (n - ) ) n - = 6n - Evaluate the expression. ) Find the first term, the common difference, and give a recursive formula for the arithmetic sequence. 4) 6th term is -0; th term is -46 ) 6 Find the indicated term using the given information. 4) a 48 = - 7, a 6 = - 7 ; a Expand the expression using the Binomial Theorem. 4) (x - )

4 ) (4x + ) 6) (ax + by) 4

5 Testname: CH 9 AND 0 AND AND EXAM REVIEW ) v = -0i - j ) ) 74-4) u = i + 4 j ) 46 6) 88. 7) D 8) Orthogonal 9) vertex: (-, -) focus: (, -) directrix: x = -8 0) (y - 6) = -4(x - 8) ) vertex: (, -) focus: (, -) directrix: x = 0 ) center at (-, -) foci at (- +, -), (- -, -) vertices at (-9, -), (, -) ) (x - 9) + y 64 = center: (9, 0); foci: (9, 7), (9, - 7); vertices:(9, 8), (9, -8) 4) (x - ) 64 + (y - ) 8 =

6 Testname: CH 9 AND 0 AND AND EXAM REVIEW ) (x + ) + (y + ) 6 = 6) 7) 4(x + ) - x 9 - y 7 = (y + ) 9 = 8) center at (, -) transverse axis is parallel to x-axis vertices at (-, -) and (8, -) 9) foci at ( - 4, -) and ( + 4, -) asymptotes of y + = - (x - ) and y + = (x - ) 0) x x - ) + x - ) ) (x - ) + - x + 4 x + - x x + 8 x + (x + ) + 6x - (x + ) 6

7 Testname: CH 9 AND 0 AND AND EXAM REVIEW 4) (9, ) ) x = 4, y = 4; x = -, y = 9 or (4, 4), (-, 9) 6) x = -4, y = -8 or (-4, -8) 7) x =, y = ; x = -, y = ; x =, y = -; x = -, y = - or (, ), (-, ), (, -), (-, -) 8) x =, y = ; x =, y = -; x = -, y = ; x = -, y = - or (, ), (, -), (-, ), (-, -) 9) x =, y = ; x = -, y = - 0) or,, -, - 0 ) s= 6, s= 8, s= 0, s4= 4, s= 4 ) c = - 4, c =, c = - 48, c 4 = 6, c = - 80 ) a =, a =, a = -, a4 = - 4) (n + ) ) 0 k k + k = 6) 7) 80 8) an = n - n 9) an = 8 + n; a = 4 40) 4) a = 0, d = -4, an = an ) - 4) 84,97 44) 040 7

8 Testname: CH 9 AND 0 AND AND EXAM REVIEW 4) a = 9 6 ; a n = 9 46) an = -4 n- 47) 6 n- 48) Converges; 4 49) Diverges 0) First we show that the statement is true when n =. For n =, we get = () (() - ) =. This is a true statement and Condition I is satisfied. Next, we assume the statement holds for some k. That is, (k - ) = k (k - ) is true for some positive integer k. We need to show that the statement holds for k +. That is, we need to show that ((k + ) - ) = k + ((k + ) - ). So we assume that (k - ) = k (k - ) is true and add the next term, (k + ) -, to both sides of the equation (k - ) + (k + ) - = k (k - ) + (k + ) - = [k(k - ) + 0(k + ) - 6] = (k - k + 0k + 0-6) = (k + 9k + 4) = (k + )(k + 4) = k + (k + - ) = k + ((k + ) - ) Condition II is satisfied. As a result, the statement is true for all natural numbers n. 8

9 Testname: CH 9 AND 0 AND AND EXAM REVIEW ) First, we show that the statement is true when n =. For n =, we get (or 6 [() - ] ) = 6() - = =. This is a true statement and Condition I is satisfied. Next, we assume the statement holds for some k. That is, k - = 6k - is true for some positive integer k. We need to show that the statement holds for k +. That is, we need to show that k = 6k + -. So we assume that k - = 6k - is true and add the next term, 6 k, to both sides of the equation k k = 6k k = 6k k = 6 6k - = 6k + -. Condition II is satisfied. As a result, the statement is true for all natural numbers n. ) 7,08 ) 0 4) x - 0x x - 80x + 80x - ) 04x + 840x x + 40x + 60x + 4 6) a x + a 4 bx 4 y + 0a b x y + 0a b x y + ab 4 xy 4 + b y 9