( )( ) ( ) MTH 95 Practice Test 1 Key = 1+ x = f x. g. ( ) ( ) The only zero of f is 7 2. The only solution to g( x ) = 4 is 2.

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1 Mr. Simonds MTH 95 Class MTH 95 Practice Test 1 Key 1. a. g ( ) ( ) + 4( ) 4 1 c. f ( x) x 14 e f g 1+ g. f i. g ( 4) ( 4) + 4( 4) k. g( x) x x x 4x+ 4 0 x b. f x 14 7 x 1 d. g( x+ ) ( x+ ) + 4( x+ ) f. f ( 4+ ) f ( 6) x+ x+ + 4x+ x + 4x x+ x 4x 4+ 4x+ x f ( ) f ( ) 6+ 7 ( 1) j. 0 f x x x Te only solution to g 4 is.. a. Te solutions are 6 and. b. Te only solution is 0. Te only zero of f is 7. c. Te solution set is (,0]. d. Te solution set is ( 5, ) (, ) e. m( x) + 7 m( x) m( x) 7 Te only solution is 9. m x 7. Practice Test 1 Key 1

2 Mr. Simonds MTH 95 Class Spring Term Te slope of te line (from rise over run ) is. Plugging m and into y mx+ b we get: te coordinates of te point (, 7) 7 + b 1 b x y So it looks like te equation of te line is y x 1 So, te function formula is f ( x) x 1.?. Cecking te point ( 5, 16) we ave: 4. Note: Wat follows is one way to come up wit te correct answer. You don t need to sow te same work, you just need to figure out te answer (to get credit). Te function value is determined from te y-coordinate. Solving te equation for y we get y x 9 +. So te function formula is f ( x) x+ 9 g a. g x b. 15 x x 6 x Te solution is. 5. Te grap of te function is a parabola so te domain is (, ). Clearly te y-coordinates are never negative, so te range is [ 0, ). (You can also see tat is te range if you picture te grap of te parabola in your ead or sketc it out on paper.) 6. a. 4 b. c. d. e. 5 and f. ( 5, ) g. 1 and i. (, ) j. (, ) (, 1) ( 1,4] k. (, ) (f is also positive at 1.) 7. a. g ( 4) ( 4) 6 l. (, ) (, ) b. f ( 4) + g( 4) ( 4) + [ 4] 14 + ( ) 1 P ractice Test 1 Key

3 Mr. Simonds MTH 95 Class 7. c. g( 5+ ) g( ) 6 e. g( + ) ( + ) d. ( 5) f ( 5 5) f ( 0) 0 f. g( ) g ( + ) [ ] + 0 1; so long as doesn't equal zero 5, and te axis of symmetry is te line x 5. Te parabola as no x-intercepts. b. Te vertex is te point ( 5, ) and te axis of symmetry is te line x 5. Te parabola as two x-intercepts. c. Te vertex is te point ( 5, ) and te axis of symmetry is te line x 5. Te parabola as no x-intercepts. d. Te vertex is te point ( 0,4 ) and te axis of symmetry is te line x 0. Te parabola as no x-intercepts. e. Te vertex is te point ( 4,0) and te axis of symmetry is te line x 4. Te parabola as one x-intercept. b f. Please note tat is OK if you use te formula x to start finding te vertex in tis a problem. I coose, toug, to complete te square.. a. Te vertex is te point Te vertex is te point ( 4, 4) as two x-intercepts. y x x+ 1 y x x y 4 4 and te axis of symmetry is te line x 4. Te parabola Practice Test 1 Key

4 Mr. Simonds MTH 95 Class Spring Term a. 1 y x x y x 1 x+ 6 6 y 6 6 c. 1 y x + 5 x y x + 5 x y x+ b. y x + x 16 5 y x x y Please note: On your actual test tese type problems will be on te calculator portion of te test so tat you can grap and ceck your answer. Figure From te vertex we ave y a ( 0,0 ) we get: + 6 for some unknown number a. Using te point 0 a a a a So te vertex form of te equation is y x+ 6. Expanding we get: y ( x+ ) 6 y ( x + 4x+ 4) 6 y x + 6x+ 6 6 y x + 6 x So te standard form of te equation of te parabola is + 6. y x x 4 P ractice Test 1 Key

5 Mr. Simonds MTH 95 Class Figure 4 From te vertex we ave y a ( 6, ) we get: + for some unknown number a. Using te point a 6 + a a a So te vertex form of te equation is y x +. Expanding we get: y ( x ) + y ( x 6x+ 9) + y x + 4x 6+ y x + 4x So te standard form of te equation of te parabola is + 4. y x x 11. Te domain is (,1) [, ) and te range is (,] 1. Table 1 Expression Exact Value Decimal Value 7 1 ( 5 5 ) ( 5 65) / y y b. Te solutions are 1,, and a. Practice Test 1 Key 5

6 Mr. Simonds MTH 95 Class Spring Term 010 c. Te points are (.5,.51), ( 0.5,1.96), and d. Te domain is (, ) and te range is (,0.65] 14. Te domain of g is (,] and te range is (,1] 1.77, (See Figure 1K.). (See figures K and 6.) Figure 1K Figure K Figure 6: y g( x) 15. a. Te maximum eigt reaced by te pebble was ft. (See Figure K.) b. Te pebble was 7.96 ft above te ground 1. seconds after it was tossed. (See figures 4K and 5K; you can simply type 1. and tap OK after selecting te trace option.) c. Te pebble reaced te ground after about.57 seconds. (See figures 6K-K.) d. Te pebble was exactly 15 feet above te ground approximately 0.5 seconds and.15 seconds after it was tossed. (See figures 9K-1K) Figure K Figure 4K Figure 5K Figure 6K Figure 7K Figure K Figure 9K Figure 10K Figure 11K Figure 1K 6 P ractice Test 1 Key