0 1 2 3 4 5 ( ) 3 6 12 24 48 96 ( )

Solutions
Name _______________________________
Block ____ Date __________
Algebra: R.2 Functions
Bell Work: What is the domain and range for each function?
y
a.
b.
c.
y
y
(3,5)
5
x
3
Domain: 𝟎 ≀ 𝒙 ≀ πŸ‘
Range: 𝟎 ≀ π’š ≀ πŸ“
(5, 3)
x
x
Domain: 𝒙 ∈ ℝ
Range: π’š ≀ πŸ“
Domain: 𝒙 ≀ πŸ“
Range: π’š ≀ πŸ‘
𝒉(𝒙)
1. Use functions f, g and h to answer the following questions.
14
𝒇
y
12
10
8
Multiply by 3 and
then subtract 2
6
4
2
𝒙
βˆ’12
0 1
2
3
4
βˆ’10
βˆ’8
βˆ’6
βˆ’4
π’ˆ(𝒙) 3 6 12 24 48 96
If 𝑔(π‘₯) = 48, then what is π‘₯?
h.
𝒇(𝒙) = πŸ‘πŸ‘ βˆ’ 𝟐
c.
Classify f
i.
d.
𝑳𝑳𝑳𝑳𝑳𝑳
Write an equation in vertex form for h
j.
e.
𝒉(𝒙) = (𝒙 + πŸ’)𝟐 βˆ’ πŸ”
What is β„Ž(βˆ’8)?
k.
Classify h
Classify g
l.
𝑸𝑸𝑸𝑸𝑸𝑸𝑸𝑸𝑸
f.
π‘₯=πŸ’
= 𝟏𝟏
𝑬𝒙𝒙𝒙𝒙𝒙𝒙𝒙𝒙𝒙𝒙
6
βˆ’6
Write an equation for f
b.
4
βˆ’4
g.
3(βˆ’4) βˆ’ 2 = βˆ’πŸπŸ
2
βˆ’2
What is 𝑓(βˆ’4)?
a.
βˆ’2
5
x
If β„Ž(π‘₯) = βˆ’2, then what is π‘₯?
π‘₯ = βˆ’πŸ”
What is 𝑔(3)?
π‘₯ = βˆ’πŸ
= 𝟐𝟐
If 𝑓 (π‘₯) = 16, then what is π‘₯? 3π‘₯ βˆ’ 2 = 16
Write an equation for g
π’ˆ(𝒙) = πŸ‘(𝟐)𝒙
3π‘₯ = 18
𝒙=πŸ”
2. Aura thinks the solution to the systems of equations below is (βˆ’2, 4) while Edison thinks the solution is
(2, βˆ’2). Teresito thinks they are both wrong. Who is correct? Solve the system to Justify your answer.
3(βˆ’2) + 2(4) = 2
5(βˆ’2) βˆ’ 12 = 4 
3π‘₯ + 2𝑦 = 2
5π‘₯ βˆ’ 12 = 𝑦
Edison is Correct
3(2) + 2(βˆ’2) = 2
5(2) βˆ’ 12 = βˆ’2 
3π‘₯ + 2(5π‘₯ βˆ’ 12) = 2
3π‘₯ + 10π‘₯ βˆ’ 24 = 2
13π‘₯ = 26
𝒙=𝟐
3. Graph the function β„Ž(π‘₯ ) = βˆ’βˆš3 βˆ’ π‘₯ .
Use inequalities to describe the domain and range.
5(2) βˆ’ 12 = 𝑦
π’š = βˆ’πŸ
y
8
6
4
Domain: 𝒙 ≀ πŸ‘
Range: π’š ≀ 𝟎
2
x
βˆ’8
βˆ’6
βˆ’4
βˆ’2
2
4
6
8
βˆ’2
βˆ’4
βˆ’6
βˆ’8
4. If 𝑓(π‘₯ ) =
βˆ’5
π‘₯+2
and g(x) = (x βˆ’ 2)3, find each output value below (if possible). If it is not possible,
explain why not.
a. f(βˆ’2) Not Possible, you can’t divide by 0
b. g(βˆ’1) = βˆ’πŸπŸ
c. g(4)
= πŸ–
d. f(βˆ’7) + g(1) = 1 + (βˆ’1) = 𝟎
e. f(3) βˆ’ g(2)
= βˆ’1 + 0 = βˆ’πŸ
5. Write and solve a system of equations to solve the following problem. Be sure to define your
D = the number of dimes
variables and answer the question at the end.
Q = the number of quarters
Jessica has 147 coins that are all dimes and quarters. The number of quarters
is 6 fewer than twice the number of dimes. What is the value of her coins?
𝑫 + 𝑸 = 𝟏𝟏𝟏
𝑸 = 𝟐𝟐 βˆ’ πŸ”
𝐷 + 2𝐷 βˆ’ 6 = 147
3𝐷 = 153
𝐷 = 51
𝑄 = 2(51) βˆ’ 6
𝑄 = 96
Value = 51(. 10) + 96(.25)
= $𝟐𝟐. 𝟏𝟏
PARCC Prep
6.
2
7.
y
8
x
2
0
y
0
-3
6
4
2
x
βˆ’8
βˆ’6
βˆ’4
2
βˆ’2
4
6
8
βˆ’2
βˆ’4
βˆ’6
βˆ’8
8.
𝑏(0) = 4(2)0 = 4
βˆ†π‘
𝑏(5) = 4(2)5 = 128
Δ𝑏 124
𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
=
= 24.8 π‘šπ‘šπ‘šπ‘šπ‘šπ‘š
5
Δ𝑑
9. Use a graph in calculator to solve the system of equations.
Write solution(s) in (x, y) form. Make a sketch of the graph
with the solutions labeled. (hint: enlarge the window!)
y = x2 βˆ’ x + 12
y = 2x2 + 3x + 7
y
(βˆ’5, 42)
(1, 12)
x
10. Write an equation for f(x).
𝒇(𝒙) = βˆ’(𝒙 βˆ’ 𝟏)𝟐 + πŸ’
11. Solve each equation or inequality below, if possible by using graphs.
a.
βˆ’(x βˆ’ 3)(x + 1) = 4
y
1
𝒙=𝟏
12.
b. |π‘₯ βˆ’ 3| > 4
c. (x βˆ’ 5)3 = 8
y
x
βˆ’1
y
x
x
7
7
𝒙 < βˆ’πŸ or 𝒙 > πŸ•
𝒙=πŸ•
Graph the parent function 𝑦 = |π‘₯| (dotted) and then graph
𝒇(𝒙) = |𝒙 βˆ’ πŸ‘| βˆ’ 𝟏 (solid) by transforming the parent.
Label the vertex, intercepts and roots on the graph of f.
y
8
6
4
2
x
βˆ’8
βˆ’6
βˆ’4
βˆ’2
2
4
6
8
βˆ’2
(3, βˆ’1)
Vertex
Roots(x-intercepts):
𝒙 = 𝟐, 𝒙 = πŸ’
βˆ’4
π’š intercept: π’š = 𝟐
βˆ’6
βˆ’8