University:
Santa Ana College
Course:
MATH 170 | Pre-Calculus Mathematics
Academic year:

2019
Views:

341

Pages:

7
Author:

encadenasyg0c

1) Solve each problem algebraically: a) 2 𝑥 2 − 8 𝑥 + 5 = 0 2x 2 −8x+5=0 𝑥 = − 𝐵 ± 𝐵 2 − 4 𝐴 𝐶 2 𝐴 x= 2A −B± B 2 −4AC 𝑥 = 8 ± ( − 8 ) 2 − 4 ( 2 ) ( 5 ) 4 x= 4 8± (−8) 2 −4(2)(5) 𝑥 = 8 ± 64 − 40 4 x= 4 8± 64−40 𝑥 = 8 ± 24 4 = 8 ± 2 6 4 x= 4 8± 24 = 4 8±2 6 𝑥 = 2 ± 6 2 x=2± 2 6 or 𝑥 = 4 ± 6 2 x= 2 4± 6 b) 2 − 2 𝑥 − 5 = 3 𝑥 2−2x −5=3x 2 − 2 𝑥 = 3 𝑥 + 5 2−2x =3x+5 ( 2 − 2 𝑥 ) 2 = ( 3 𝑥 + 5 ) 2 ( 2−2x ) 2 =(3x+5) 2 2 − 2 𝑥 = 9 𝑥 2 + 30 𝑥 + 25 2−2x=9x 2 +30x+25 0 = 9 𝑥 2 + 32 𝑥 + 23 0=9x 2 +32x+23 𝑥 = − 32 ± 3 2 2 − 4 ( 9 ) ( 23 ) 2 ( 9 ) x= 2(9) −32± 32 2 −4(9)(23) 𝑥 = − 32 ± 1024 − 828 18 x= 18 −32± 1024−828 𝑥 = − 32 ± 196 18 x= 18 −32± 196 𝑥 = − 32 ± 14 18 x= 18 −32±14 𝑥 = − 32 + 14 18 𝑜 𝑟 𝑥 = − 32 − 14 18 x= 18 −32+14 orx= 18 −32−14 𝑥 = − 1 𝑜 𝑟 𝑥 = − 46 18 = − 23 9 x=−1orx= 18 −46 = 9 −23 2) Solve each inequality algebraically, graph the solution set on a number line, and state the solution in interval notation: a) 2 𝑥 2 + 3 𝑥 − 9 ≥ 0 2x 2 +3x−9≥0 ( 2 𝑥 − 3 ) ( 𝑥 + 3 ) ≥ 0 (2x−3)(x+3)≥0 𝑥 = 3 2 𝑜 𝑟 𝑥 = − 3 x= 2 3 orx=−3 Interval notation: ( − ∞ , − 3 ] ∪ [ 3 2 , ∞ ) (−∞,−3]∪[ 2 3 ,∞) b) 1 2 𝑥 + 5 ≥ 2 2 1 x+5≥2 1 2 𝑥 ≥ − 3 2 1 x≥−3 𝑥 ≥ − 6 x≥−6 or 1 2 𝑥 + 5 ≤ − 2 2 1 x+5≤−2 1 2 𝑥 ≤ − 7 2 1 x≤−7 𝑥 ≤ − 14 x≤−14 Interval notation: ( − ∞ , − 14 ] ∪ [ − 6 , ∞ ) (−∞,−14]∪[−6,∞) 3) Find the center and radius of the circle and then sketch: 𝑥 2 + 𝑦 2 − 2 𝑥 + 4 𝑦 = 4 x 2 +y 2 −2x+4y=4 𝑥 2 − 2 𝑥 + 𝑦 2 + 4 𝑦 = 4 x 2 −2x+y 2 +4y=4 ( 𝑥 − 1 ) 2 + ( 𝑦 + 2 ) 2 = 9 (x−1) 2 +(y+2) 2 =9 Center: ( 1 , − 2 ) (1,−2), Radius: 3 3 4) Lines: a) Find the equation of the line that passes through ( 0 , − 1 ) (0,−1) and ( 5 , 9 ) (5,9) 𝑦 − 𝑦 1 = 𝑚 ( 𝑥 − 𝑥 1 ) y−y 1 =m(x−x 1 ) 𝑦 − ( − 1 ) = 2 ( 𝑥 − 0 ) y−(−1)=2(x−0) 𝑦 + 1 = 2 𝑥 y+1=2x 𝑦 = 2 𝑥 − 1 y=2x−1 b) Find the equation of the line that contains ( − 3 , − 4 ) (−3,−4) and is perpendicular to 5 𝑥 + 6 𝑦 + 7 = 0 5x+6y+7=0. Slope of given line: − 5 6 6 −5 Perpendicular slope: 6 5 5 6 𝑦 + 4 = 6 5 ( 𝑥 + 3 ) y+4= 5 6 (x+3) c) Find the equation of the line whose graph is given: 𝑦 = 3 2 𝑥 − 2 y= 2 3 x−2 5) Solve graphically to two decimal places: 𝑥 4 − 8 𝑥 2 = − 5 x 4 −8x 2 =−5 𝑥 ≈ ± 2.70 x≈±2.70 𝑥 ≈ ± 0.83 x≈±0.83 6) Given function f(x) find each: a) Domain: [ 7 , ∞ ) [7,∞) b) Range: [ − 3 , 4 ] [−3,4] c) 𝑓 ( 0 ) = 4 f(0)=4 d) 𝑓 ( − 2 ) = 2 f(−2)=2 e) Find all x values where 𝑓 ( 𝑥 ) = 4 f(x)=4: 𝑥 = 0 x=0 and 𝑥 = − 7 x=−7 7) Given 𝑓 ( 𝑥 ) = 4 𝑥 2 − 3 𝑥 2 f(x)=4x 2 −3x 2 find each: a) 𝑓 ( 2 ) = 4 f(2)=4 b) 𝑓 ( − 4 ) = − 64 f(−4)=−64 c) 𝑓 ( 𝑎 + 1 ) = 3 𝑎 2 − 2 𝑎 + 1 f(a+1)=3a 2 −2a+1 d) 𝑓 ( 𝑎 + ℎ ) − 𝑓 ( 𝑎 ) ℎ = 4 ℎ − 6 𝑎 + 3 ℎ h f(a+h)−f(a) =4h−6a+3h 8) Graph the function and find the intervals: a) Increasing: [ 5.25 , ∞ ) [5.25,∞) b) Decreasing: ( − ∞ , 5.25 ] (−∞,5.25] c) Average rate of change: − 51 −51 9) Given f(x), g(x), h(x): a) ( 𝑓 ∘ 𝑔 ) ( 𝑥 ) = 9 𝑥 2 + 12 𝑥 − 3 𝑥 + 1 (f∘g)(x)= x+1 9x 2 +12x−3 b) ( ℎ ∘ 𝑔 ) ( 𝑥 ) = − 15 𝑥 − 7 3 𝑥 + 3 (h∘g)(x)= 3x+3 −15x−7 c) Inverse of 𝑔 ( 𝑥 ) g(x): 𝑔 − 1 ( 𝑥 ) = 𝑥 − 2 3 g −1 (x)= 3 x−2 d) Inverse of ℎ ( 𝑥 ) h(x): ℎ − 1 ( 𝑥 ) = 3 − 𝑥 − 𝑥 + 5 h −1 (x)= −x+5 3−x

Math 170 Practice Test 2 Solutions

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