Name ____Key_____________________________ Date ___________ Period _______ Pre-Calculus Unit 1 Practice Test Complete the problems below and show your work. Target 1A: I can identify functions from data tables, graphs, and descriptions of set relations. 1. Does the graph below represent a function? Explain. y yes; each π₯-value has one π¦-value and it passes the vertical line test x 2. Does the table represent a function? Explain why or why not. π 4 1 −3 8 1 π 2 6 3 8 9 No; the π₯-value of 1 has two different π¦values. 3. If π(π₯) = −π₯ 2 + 2 evaluate a. π(2) −(2)2 + 2 = −4 + 2 = −2 b. π(−1) −(−1)2 + 2 = −1 + 2 = 1 c. π(3) − π(1) π(3) = −(3)2 + 2 = −7 and π(1) = −(1)2 + 2 = 1 so π(3) − π(1) = −7 − 2 = −8 Target 1B: I can describe a set of numbers in a variety of ways. For each of the following, fill in the missing type of interval or graph. Describe the interval as bounded, unbounded, open, closed, half-open. 4. Interval Graph (3, 7] Inequality ← ] 0 1 2 3 4 5 6 7 Description π»πππ − ππππ πππ’ππππ 3<π₯≤7 (3, ∞) 5. Interval Graph π₯>3 Inequality ( 0 1 2 3 4 5 6 7 Description πππππ’ππππ πππ ππππ (−∞, 9] 6. Interval Inequality π₯≤9 Graph 0 9 π’ππππ’ππππ ππππ ππ Description__________________ 7. Describe the set of numbers using interval notation. π₯ ≥ 5 or π₯ < 11 [5, ∞)ππ (−∞, 11) (−∞, 11)ππ [5, ∞) 8. Describe the set of numbers using set-builder notation. {−9, −8, −7, −6, −5, … } {π₯|π₯ ≥ −9, π₯ ∈ β€} 9. Describe the domain and range of π¦ = √π₯ + 3 in interval notation. π·πππππ: [−3, ∞) Range: [0, ∞) 10. Use the graph below to find the domain and range. y 10 8 π·πππππ: (−2, −1]π(0, 3] π ππππ: (−7, 6.2] 6 4 2 –10 –8 –6 –4 –2 –2 2 4 6 8 10 x –4 –6 –8 –10 11. Find the domain and range of the relation {(−2, 4), (3, 5), (4, −2), (3, 8)} and explain if it determines a function. π·πππππ: {−2, 3, 4} π ππππ: {−2, 4, 5, 8} Not a function because 3 has two different π¦-values Target 1C: I can define, interpret, and use piecewise functions in function notation and as a graph. 3π₯ − 1 ππ π₯ < −3 13. Graph π(π₯) = {π₯ + 4 ππ − 3 ≤ π₯ < 2 −2 ππ π₯ > 2 2π₯ + 1 ππ π₯ < 0 12. Graph π(π₯) = { 4π₯ ππ π₯ ≥ 0 9 y 8 7 6 5 4 3 2 1 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 14. Graph π(π₯) = { 9 y 8 7 6 5 4 3 2 x 1 -1 1 2 3 4 5 6 7 8 9 -9 -8 -7 -6 -5 -4 -3 -2 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 x 1 2 3 4 5 6 7 8 9 π₯ 2 ππ π₯ < 0 5π₯ ππ π₯ ≥ 0 9 y 8 7 6 5 4 3 2 1 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 x 1 2 3 4 5 6 7 8 9 15. Write a piecewise function for the graph below. 9 8 7 6 5 4 3 2 1 -9 -8 -7 -6 -5 -4 -3 -2 -1-1 -2 -3 -4 -5 -6 -7 -8 -9 y −π₯ + 3, ππ π₯ < 0 π(π₯) = { −π₯ 2 , ππ π₯ ≥ 0 x 1 2 3 4 5 6 7 8 9 16. Rewrite the function in the previous question so that the function would be continuous. −π₯, ππ π₯ < 0 π(π₯) = { 2 −π₯ , ππ π₯ ≥ 0 17. Write a piecewise function for the graph below. 9 8 7 6 5 4 3 2 1 y -9 -8 -7 -6 -5 -4 -3 -2 -1-1 -2 -3 -4 -5 -6 -7 -8 -9 π₯ 2 , ππ π₯ ≤ 0 π(π₯) = { π₯ − 2, ππ π₯ > 0 x 1 2 3 4 5 6 7 8 9 18. Rewrite the function in question 17 so that the function would be continuous. π₯ 2 , ππ π₯ ≤ 0 π(π₯) = { π₯ − 2, ππ π₯ > 0 Target 1D: I can determine the average rate of change for a function as well as identify increasing and decreasing functions and intervals. 19. For which interval(s) is the function π¦ = 2π₯ 3 − 8π₯ + 5 increasing and decreasing? πΌππ.: (−∞, −1)π(1.2, ∞) π·ππ.: (−1, 1.2) 20. Find the extrema for π(π₯) = −3π₯ 3 + 8π₯ 2 + 10π₯ − 9 name the specific type of extrema. πΏππππ πππ: (−0.49, −11.63) πΏππππ πππ₯: (2.27, 19.83) 21. Graph the function π¦ = π₯ 4 + 2π₯ 3 + 3π₯ on your calculator. Find the x-value of any extrema to the nearest hundredth and describe what type of extrema it is. πΊπππππ πππ π₯ = −1.75 22. Find the average rate of change for π(π₯) = π₯ 3 − π₯ 2 on the following intervals. a. [0, 4] b. [−4, −3] π(0) = 0 π(4) = 43 − 42 = 48 48 − 0 = 12 4−0 π(−4) = (−4)3 − (−4)2 = −80 π(−3) = (−3)3 − (−3)2 = −36 −36 − −80 44 = = 44 −3 − −4 1 23. Find the average rate of change for π(π₯) = π₯ 2 + π₯ on the following intervals. a. [1, 3] π(1) = 12 + 1 = 2 π(3) = 32 + 3 = 9 + 3 = 12 12 − 2 10 = =5 3−1 2 b. [−4, −1] π(−4) = (−4)2 − 4 = 12 π(−1) = (−1)2 − 1 = 1 − 1 = 0 0 − 12 12 =− = −4 −1 − (−4) 3 c. [π, π + β] π(π) = π2 + π π(π + β) = (π + β)2 + π + β = π2 + 2πβ + β2 + π + β (π2 + 2πβ + β2 + π + β) − (π2 + π) π2 + 2πβ + β2 + π + β − π2 − π 2πβ + β2 + β = = = 2π + β + 1 π+β−π β β 24. Find the average rate of change for the graph below on the interval [−2,0]. y 9 8 7 2 6 5 4 3 2 1 -5 -4 -3 -2 -1 x 1 -1 2 3