1. PARCC Released Items 2016 #1 to #16

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
average rate of descent = change in feet
change in sec.
= 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)