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