Name: Sept 21, 2017 Page 1 of 1

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1 MATH (Kunkle), Eam pts, 75 minutes No notes, books, electronic devices, or outside materials of an kind. Read eac problem carefull and simplif our answers. Name: Sept 21, 2017 Page 1 of 1 Unless oterwise indicated, supporting work will be required on ever problem wort more tan 2 points. 1(25 pts). Solve te inequalities. Write te solution set in interval notation. a b c (7 pts). Find an equation of te circle tat is centered at (3, 1) and passing troug te point ( 2, 4). 3(5 pts). Find an equation of te line passing troug te points ( 3,7) and (1,2). 4(7 pts). Find an equation of te line passing troug te point ( 3,7) and perpendicular to 3+2 = 5. 5(10 pts). Use factorization, binomial epansion, or rationalization to find and cancel te common factor in eac epression. a b. 12 3(+2) 2 6(16 pts). Grap te function. Find all intercepts and asmptotes, if an, and include tem in our grap. State te domain and range of te functions in parts b and d onl. a. f() = b. () = 3 c. g() = 4/3 d. k() = 4/3 16 7(6pts). Rewriteteepression +1 2 witoutabsolutevaluesmbolsif 1 < 2. 8(14 pts). Let f() = 1+ and g() = Find eac of te following functions and teir domain. Simplif our answers and epress te domains in interval form. ( ) ( ) a. f g () b. g f () 9(6 pts). Find all - and -intercepts of te grap of p() = Label our answers so I can tell wic is wic. { 2 1 if 0 +5 if > Let q() = Work in te space below and label our answers to te following questions. a(5 pts). Rewrite q() in standard form. b(6 pts). Find te coordinates of te verte and all intercepts. c(2 pts). Give te range of q() in interval notation.

2 (Kunkle) Eam 1 9/21/2017 Solutions 1a.(Source: ) First solve te related equation: 4 7 = 17 means 4 7 = 17 4 = 24 = = 17 4 = 10 = 5/2 Tese solutions divide te -ais into tree intervals. Test te inequalit on eac. Here s wat I found wen I did tis: : T F T : Including te endpoints (since te inequalit is true at tese), te solution set is (, 5/2] [6, ). 1b.(Source: ,1.1.43) Factor and make a sign cart: 3 16 = ( 2 16) = ( 4)(+4) 0. 4 : : : ( 4)(+4) : : Solution set is [ 4,0] [4, ). 1c.(Source: ,more.1i) Get a zero on one side, factor te oter, and ten make a sign cart: (+3) 2 0 = = : : : DNE +3 : 4 3 Solution set is (, 4] ( 3, ) 2.(Source: ) Te radius of te circle is te distance from (3, 1) to ( 2, 4), or = 34. Te equation of te circle is ( 3) 2 +( +1) 2 = (Source: ) Slope = = = 5 4. In point-slope form, te line can be written 7 = 5 4 (+3), or 2 = 5 4 ( 1). 4.(Source: ) First find te slope of = 5 b converting to slope-intercept form: 2 = 3+5 = = Tis line as slope 3, so te desired line must ave 2 slope 2 3. Its equation is 7 = 2 3 (+3).

3 (Kunkle) Eam 1 9/21/2017 Solutions 5a.(Source: ) Rationalize te denominator: ( 3) ( +1 2) ( +1+2) ( +1+2) = ( 3)( +1+2) = ( 3)( +1+2) 3 = ( 3)( +1+2) +1 4 = b.(Source: ) Epand te binomial: 12 3(+2) 2 = 12 3(2 +4+4) = = ( 3 12) = ) = (Source: 2.2.more.1o) To obtain te grap of = 3, reflect = across te -ais, and sift te result rigt 3 units. To see tis, it elps to find te domain and intercepts of 3 algebraicall. Tis function requires 0 3, or 3, so its domain is (,3]. To find te -intercept, set = 0 and calculate = 3 0 = 3. To find te -intercept, set = 0 and solve for : 3 = 0 3 = =. Sifting rigt didn t cange te range of te square root: [0, ). (0, 3) (0,0) (3,0) a. = b. = 3 In quadrant 1, te grap of 4/3 looks like tat of 2, since 4/3 > 1. Tis function equals ( 3 ) 4, wic is defined for < 0 and is even. Terefore its grap is smmetric across te -ais. ( 8, 0) (8, 0) (0,0) (0, 16) c. = 4/3 d. = 4/3 16

4 (Kunkle) Eam 1 9/21/2017 Solutions Te grap of 4/3 16 is obtained b sifting tat of 4/3 downward 16 units, making te new -intercept (0, 16). To find te -intercept, set = 0 and solve 0 = 4/3 16: 4/3 = 16 = ±16 3/4 = ±( 4 16) 3 = ±2 3 = ±8. Te -intercepts are (8,0) and ( 8,0). Avoid working wit large numbers. After te step marked, I took te cube root and ten raised te result to te power 4. If ou took te 4t power before taking te cube root, ou were left aving to calculate Tedomainandrangeof 4/3 are(, )and[0, ),respectivel. Becauseoftevertical sift, te domain and range of 4/3 16 are (, ) and [ 16, ), respectivel. 7.(Source: ) +1 = { +1 if +1 0 (+1) if = { 2 if 2 0 ( 2) if 2 0 If 1 2, ten 0 +1 and 2 0, and so +1 2 = +1 ( ( 2)) = = 2 1. ( ) 8a. (Source: 2.6.more.1d,2d) f g () = 1+(2 10) = 2 9. Te domain is te solution set to = ( 3)(+3), wic we find wit a sign cart. +3 : : : : 3 3 Hence te domain of ( f g ) () is (, 3] [3, ). 8b. ( g f ) () = ( 1+) 2 10, wic, wen defined, simplifies to 9. For tis function to be defined, te radical requires tat 1+ 0, or 1. Hence te domain is [ 1, ). 9. (Source: ,7-8) Find te -intercept b evaluating p() at = 0. Since te first case applies, p(0) = = 1, and terefore te -intercept is (0, 1). Some students said tat te grap ad two -intercepts, an impossibilit along te grap of a function. Wen we set 2 1 = 0 and +5 = 0 we find tat te onl possible -intercepts are = 1/2 and = 5. Te first is an intercept, since = 1/2 is on tat part of te domain were p() is given b 2 1. But since = 5 0, we do not use +5 to calculate p( 5). Onl 1/2 is an -intercept.

5 (Kunkle) Eam 1 9/21/2017 Solutions 10a. (Source: ,more.1j) q() = 2( 2 10) 32 = 2( ) = 2( 5) b. Verte is (5,18). Wen = 0, calculate = 32, so te -intercept is (0, 32). Set = 0 and solve for, using te standard form. 2( 5) = 0 2( 5) 2 = 18 ( 5) 2 = 9 5 = ±3 = 5±3 = 8 or 2 so te -intercepts are (8,0) and (2,0). We could also ave solved q() = 0 b factoring: = 2( 8)( 2). 10c. Te grap of q opens downward, so 18 is its maimum value. Te range is (,18]. You were not required to sketc te grap of q(), but ere s wat it looks like: (5,18) (2,0) (8,0) (0,-32)