1. PARCC Released Items 2016 #1 to #30

2.     (r – 3)2 = A Place r on the left side. (r – 3)2 = A
Isolate r: Divide both sides by 
 
(r – 3)2 = A  
(r – 3)2 = A 
Undo the square by taking the square root.
r – 3 = A  
Isolate r: Add 3 to both sides
r = A  

3. 7 7
Factor and use the Zero Product Property:
x2 – 49 is a difference of squares
(x + 7)(x + 7)(x – 7) = 0
x2 – 49 = (x + 7)(x – 7)
x + 7 = 0 or x – 7 = 0
To solve, each factor can equal zero.
– 7 –
x = – 7 or x = 7

4.    C. (0, 0) is NOT on the line (0, -1) and (2, 3)
are on the graph.
Test (15, 29): y = 2x – 1
D. Test (-0.5, -2): -2 = 2(-0.5) – 1
29 = 2(15) – 1
-2 = 1 – 1 
29 = 29 
Test (0.3, -0.4): = 2(0.3) – 1.0
B. Test (2000, 1999)
-0.4 = 0.6 – 1.0 
y = 2x – 1
1999 = 2(2000) – 1
Test (0.5, 0): = 2(0.5) – 1
1999 = False
0 = 1 – 1 
E. Test ( ¼ , -½ ): ½ = 2/1( ¼ ) – 1  
-½ = ½ – 2/2  
Test (4/5 , 3/5 ): /5 = 2/1(4/5 ) – 1
3/5 = 8/5 – 5/5 

5. 1
2x(x2 – 2) – 1(x2 – 2) – x(x2 – x – 2)
2x3 – 4x – 1x – x3 + x2 + 2x 2
2x3 – 1x2 – 4x + 2
1x3 + 1x2 + 2x 2
1x –2x + 2
ax3 + bx2 + cx + d

6.  dotted line solution: area shaded twice (pink area) solid line
Solve for y:
2x  y ≥ 1
x + y  3
Solve for y:
x x
2x 2x
y ≥ 2x + 1
Division by 1
Inequality symbol reverses
y  1x + 3 1
1 1 1
y ≤ 2x  1 1
 shade above the line
≤ shade below the line

7. 12
= change in feet = f( ½) – f(0)
change in sec ½  0
average rate of ascent
= 6 – 0 feet
½  sec
= 6
= 6/1  1/2
= 6/1  2/1
= 12 feet per second
t = 0 when she jumps into air
= 16(02) + 20(0)
= 16(0) + 0 =
= 0 feet
t = ½ when she catches the bone
= 16(½)2 + 20(½)
=  16/1( ¼ ) + 10
= 
= 6 feet

8.   Equation for a parabola: f(x) = (x – h)2 + k
with a vertex of (h, k)
f(x) = (x – h)2 + k 
f(x) = (x – 2)2 + 1
vertex = (2, 1)
2 right and 1 up from (0, 0)
Suppose that f(x) = x2 with a vertex of (0, 0)
f(x + 3) = (x + 3)2
f(x  -3) = (x  -3)2 + 0
f(x  h) = (x  h)2 + k 
vertex = (3, 0)
3 left from (0, 0)
horizontal shift 3 units to left

9.   + ( ) = 0; 0 is rational  = = 2; 2 is rational4 2 
 = (irrational number) 3 6
If you have one irr. #, you need a second
irr. # to make a perfect square under the
radical.
 +  = 2 (still irrational)
You need at least 1 irrational number to get an irrational sum. 20 
 = (still irrational) 2 10 
You need two opposite irr. numbers for a rat. sum.
If there are no irr. #, the ans. is rat.

10. To be a function, each x-value must be paired with exactly one y-value.
This means that the y-term cannot have an even exponent or an absolute value. 
Solved directly for y. 
Solved directly for y.
Not solved directly for y. The y-term has an absolute value. 
Solved directly for y. 
Solved directly for y. 
Solved directly for y.

11.     To find time when on level ground, (Find x-intercepts)
Solve: 0.005x(x – 18) = 0
0.005x = or x – 18 = 0
0.005 
x = or x = 18
The crown is in the middle (average) of the
times the road is on level ground: 
(0 + 18)  2 = 18  2 = 9
crown occurs at x = 9 
y-coordinate of crown = height above level ground
0.005x(x – 18) 
= 0.005(9)(9 – 18)
= 0.005(9)(–9)  =

12.   Type each of these equations into the TI-84:
First, we need to make the right side equal to zero.
Y1 = 2(x – 3)2
1 real solution (1 x-intercept) 
Solve 2(x + 3)2 + 3 = 0
1 1
Y1 = 2(x + 3)2 + 3
no real solution (no x-intercept)
8 8 
Solve (x  1)2  4 = 0
2 2
Y1 = (x  1)2  4
2 real solutions (2 x-intercepts)
Solve (x + 1)2 + 2 = 0
Y1 = (x + 1)2 + 2
no real solution (no x-intercept)
Solve x2 + 8x + 15 = 0
Y1 = x2 + 8x + 15
2 real solutions (2 x-intercepts)

13. Type the equation into the TI-84:
Y1 = | 6 – 3x | + 6
Use the table (2nd, Graph) to find the vertex (it looks like x = 2)
Graph the vertex at (2, 6)
Graph one point on each
side of the vertex:
Graph (1, 9) and (3, 9)
Connect the vertex to each of the points.

14. f(x) = 3x2 + 18x – 21
a = 3
f(x) = ax2 + bx + c 3 3 6
f(x) = a (x  h) k
Type into the TI-84:
Y1 = 3x2 + 18x – 21
Use the table (2nd, Graph) to find the vertex (it looks like x = 3)
Vertex = (3, 6)
Vertex = (h, k)
Place h and k into the boxes.

15. 1 7 1 7
Type into the TI-84:
Y1 = 2x2 + 4x + 5
y = a (x  h) k
Use the table (2nd, Graph) to find the vertex (it looks like x = 1)
y = a (x + h) k
Vertex = (1, 7)
Vertex = (h, k)

16. ax + c = bx + d
Bring x-terms together.
d – c
bx bx
ax  bx + c = d
Isolate the x-terms.
c c
a – b
ax – bx = d – c
Factor out x.
x(a – b) = d – c
Isolate x.
(a – b) = a – b

17. 150(1.02h)
time in hours
initial number
of bacteria
Factor
rate of growth = factor – 1
150
0.02
150(1.02)
= 1.02 – 1
= 0.02
number of bacteria when h = 1
= 150(1.02h)
= 150(1.02)1
= 150(1.02)

18.  total girls/total people = 16/31 = 0.516 total pizza/total people =
13/31 = 0.419
girls with pizza/total girls =
5/16 = 0.313
girls with pizza/total pizza =
5/13 = 0.385 

19.   This is an exponential graph, not linear. The function should
be exponential.
For the nonlinear equations, use the TI-84 and
x in place of d to see if (3, 64) and (5, 256) are on the graph:
Choice A is linear: c(d) = md + b
where m is the slope, b is the y-intercept
Press the “Y =“ key to enter the equation,
Press the key to make an exponent,
Choice B is linear: c(d) = md + b
where m is the slope, b is the y-intercept
To view the table, press “2nd” then “GRAPH”

20.   r2 = 2(18x – A)  A = 18x – 0.5r2 Place r on the left side to
keep a positive coefficient.
+ 0.5r r2
A + 0.5r2 = 18x
Isolate r.
r =
2(18x – A)  
A A
Replace 0.5 with ½ .
½ r2 = 18x – A
2(½ r2) = 2(18x – A)
Multiply by denom.
r2 = 2(18x – A) 
 

21.     2 hrs. more on Tuesday = 2 hrs. less on Wednesday # hours pay
Tues c P(c) = 15c 
Wed c – 2 P(c – 2) = 15(c – 2)
Wed. pay = 15(c – 2) 
= 15c – 30
Tuesday – Wednesday
= P(c) – P(c – 2)

22. 5(x)(x)(x) + 30(x)(x) + 35(x)6 7
= x2  6x  7 7
The product of the 2 last terms in the parentheses should be 7
(____)(1) = 7
(x)(1) = 7
(1)(x) = 7
x = 7

23.  Maximum point: (12 sec, 19.4 ft) False Max. point: (12 sec, 19.4 ft)
Reaches max. altitude after 12 sec. True
y-intercept: Projectile launched from 5 feet, not the ground. False
From t = 5 to t = 20, the function increases, then decreases. False

24. 13  False
Vertex (middle) of parabola occurs at t = 12, not t = 10 False 
hits ground when height = 0 (y-coordinate = 0); hits ground at t = False
Let a = 4 h(12 – 4) = h(12 + 4)
Let a = 4 h(12 – 12) = h( )
h(8) = h(16)
h(0) = h(24)
17.8 = True
5 = 5 True

25.
n –
= change in y = n – 5
change in x  (-3)
slope
= 0.1n – 0.5
= n – 5 10
= 1n – 5 10 =
n – B E D =
n – =
– n

26. Function Z: f(n) = 0.1n – 0.5
Domain = n can be any real number:
positive or negative; E is false B C D F
Z is linear: slope = 0.1;
y-intercept = -0.5
A. n can be any real number A is false
B. slope = 0.1; positive slope indicates an increasing function
C. slope = constant rate of change = 0.1
D (natural number) = rational number; rational number – 0.5 = rational number

27.     Discriminant Values: D is neg; no real roots
D is not perf. sq.; 2 non-rat. roots
D = 0; one real root
D is perf. sq.; 2 rat. roots
Use the discriminant b2 – 4ac
y = 1x2 – 7x + 9
(7)2 – 4(1)(9)
= 49 – 36
= 13 not a perf. sq. 
y = 1x2 – 0x + 9 
(0)2 – 4(1)(9) 
= 0 – 36
= –36 negative # 
y = 1x2 – 6x + 9
(6)2 – 4(1)(9)
= 36 – 36
= 0
y = (x – h)2 + k
y = 1x2 – 10x + 9
y = x2  1x – 1x
(10)2 – 4(1)(9) 2
y = (x – 1)2 + 8
y = x2  2x + 9
= 100 – 36
y = (x – 1)(x – 1) + 8
= 64 perf. sq.

28.  Community A: More of a spread of prices (less consistent);
Most of them less expensive
Community B:
Less of a spread of prices;
(more consistent)
Most of them more expensive 

29. Community B:
156, 157, 157, 158, 158, 159, 159, 159, 159, 160, 160, 160, 160, 160, 160, 160, 160, 161, 161, 161, 161, 161, 161, 162, 162, 163, 163, 164, 165, 165, 166, 166, 167, 167, 168, 168, 169, 170, 170, 172, 172, 174, 174, 175, 175 
45 prices: median = middle = 23rd term = 161  $161, 000
third quartile: median of terms after the circled number
middle = between 167 and 168  between 167,000 and 168,000

30. 39 39 43

31.  Begin with 256 pennies Trial 1: P(heads) = ½ , so approximately
½ of 256 = 128 pennies will return to the bag
The number of pennies in the bag keeps dividing by 2 (multiplying by ½)
This represents an exponential function.
Trial 2: P(heads) = ½ , so approximately
½ of 128 = 64 pennies will return to the bag
Trial 3: P(heads) = ½ , so approximately
½ of 64 = 32 pennies will return to the bag

32.  The function is exponential,
begins with an initial value of 256 pennies,
and the value keeps multiplying by ½
The only exponential function is B.

33.  Begin with 256 pennies; keep dividing by 2 for each trial.
Trial 1: 128 pennies
Trial 2: 64 pennies
Trial 3; 32 pennies
Trial 4: 16 pennies
Trial 5: 8 pennies
Trial 6: 4 pennies
Trial 7: 2 pennies
Trial 8: 1 penny

34.  Total number of pennies = 256
# pennies in bag + # pennies on floor = 256
f(n) + g(n) = 256

35.   40 hour week  4 preparation workers = 160 hours of preparation
40 hour week  7 packaging workers = 280 hours of packaging
preparing tomato salsa + preparing corn salsa ≤ 160
2t + 2.5c ≤ 160 
packaging tomato salsa + packaging corn salsa ≤ 280
4t + 3c ≤ 280

36.    2t + 2.5c ≤ 160 4t + 3c ≤ 280 2(20) + 2.5(45) = 152.5 true
2(45) + 2.5(30) = 165 false
2(50) + 2.5(25) = false 
2(60) + 2.5(10) = 145 true
4(60) + 3(10) = 270 true

37. 2t + 2.5c ≤ t + 3c ≤ 280
2t + 2.5(20) ≤ 160
4t + 3(20) ≤ 280
2t + 50 ≤ 160
4t + 60 ≤ 280 55
2t ≤ 110
4t ≤ 220
t ≤ 55
t ≤ 55
2.5c ≤ 160
3c ≤ 280 64
c ≤ 64
c ≤ 93.3

38. y-coordinate is positive:
sum of coordinates is greater than 2:
y  0
x + y  2
x + y  2

39. All points in the first quadrant have a
positive x-coordinate and a positive y-coordinate.
This proves that y  0.
All points in the first quadrant have a
positive x-coordinate and a positive y-coordinate.
x + y  2
x + y = positive # + positive # = positive #
positive #  0 and 0  2, so therefore positive #  2
This proves that x + y  2

40. Make an equation for Cars A and B: 70t = 65t + 45 -65t -65t
slope = 350 – 0
5 – 0
t = 9
t = 10
Car A will pass Car B 9 hours after noon, 9:00 PM.
= 350  5
Car A will pass Car C 10 hours after noon, 10:00 PM.
= 70
D = mt + b
D = 70t (y-int = 0)
– 65 = 45
+65
(0, 45) and (1, 110)
D = mt + b
+65
slope = 110 – 45
1 – 0
D = 65t + 45
= 65