UNIT 1 PRACTICE TEST - QUADRATICS
MATH 2
ANSWER KEY
#1 – 6: (2 points each)
1. Solve by factoring – show work
x2 – x = 30
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DATE______________
2. Solve by factoring – show work
9x2 – 36 = 0
x2 – x – 30 =0
9(x2 – 4) = 0
(x – 6)(x + 5) = 0
9(x – 2)(x + 2) = 0
x–6=0
x=6
x+5=0
x = -5
{-5, 6}
x+2=0
x = -2
{-2, 2}
3. Solve by factoring – show work
5x2 + 15x = 0
5x(x + 3) = 0
5x = 0
x=0
x–2=0
x=2
x+3=0
x = -3
4. Solve by factoring – show work
2x2 + x = 10
2x2 + x – 10 = 0
AC method: AC = -20
-4*5 = -20 -4 + 5 = 1
2x2 – 4x + 5x – 10 = 0
{-3, 0}
2x(x – 2) + 5(x – 2) = 0
(2x + 5)(x – 2) = 0
2x + 5 = 0
2x = -5
x = -5/2
x–2=0
x=2
{-5/2, 2}
5. A quadratic function has roots
x = {-2/3, 5}
Write this function in standard form
y = ax2 + bx + c. Show work.
x = -2/3
x=5
3x = -2
x–5=0
3x + 2 = 0
(3x + 2)(x – 5) = 0
3x2 – 15x +2x – 10 = 0
3x2 – 13x – 10 = 0
y = 3x2 – 13x - 10
6. A quadratic function has vertex (-3, 6)
and goes through the point (-1, 18). Write
this function in vertex form
y = a(x – h)2 + k. Show work.
18 = a(-1 – -3)2 + 6
18 = a(2)2 + 6
18 = 4a + 6
18 – 6 = 4a
12 = 4a
3=a 
y = 3(x + 3)2 + 6
UNIT 1 PRACTICE TEST - QUADRATICS
NAME_____________________________
MATH 2
DATE______________
2
Using the quadratic function y  3(x  1)  4 , answer the following questions: (1 point each)
7. What is the vertex of the function? __(-1, -4)____________
8. What is the axis of symmetry? ___x = -1___________
9. Does it have a minimum or maximum? _max_______ What is its value? __-4_______
10. What is the range? _____y ≤ -4 (since it’s a maximum)____________________
11. Describe all the transformations that move it from the graph of y = x2
reflected over the x-axis, narrowed by a factor of 3, left 1, down 4
The function y = x2 is translated 4 units to the right and 6 units down form a new function.
12. Write this function in vertex form (1 point)
y = (x – 4)2 – 6
13. Write this function in standard form – show work. (1 point)
y = (x – 4)2 – 6
y = (x – 4)(x – 4) – 6
y = x2 – 4x – 4x + 16 – 6
y = x2 – 8x + 10
14. Convert the following quadratic function to vertex form – show work (2 points)
y = x2 – 10x + 12
y – 12 = x2 – 10x
y – 12 + (10/2)2 = x2 – 10x + (10/2)2
y – 12 + 25 = x2 – 10x + 25
y + 13 = (x – 5)2
y = (x – 5)2 – 13
Part 2: Calculator Active:
UNIT 1 PRACTICE TEST - QUADRATICS
NAME_____________________________
MATH 2
DATE______________
2
A parabola is modeled by the function f ( x )  2x  9 x  3 . Answer the following (1 pt. each)
15. What is the axis of symmetry?
x = -b/2a = -9/(2*-2) = 9/4 or 2.25
16. Min / Max (circle one) = 7.125 or 57/8
17. What are the coordinates of the
vertex?
(2.25, 7.125)
19. What are the coordinates of the yintercept?
(0, -3)
f(2.25) = -2(2.25)2 + 9(2.25) – 3 = 7.125
18. What is the range of the function?
y ≤ 7.125
20. What are the approximate zeroes?
(round to the nearest hundredth)
{0.36, 4.14}
(make y2 = 0 and find intersection)
A projectile is launched upward from an initial height h0 feet at an initial velocity v0 feet per second. A
parabola that models this is of the form h(t )  16t 2  v0t  h0 , where h(t) is the height in feet after t
seconds. Readings for some heights and times are given in the following table:
t seconds
h(t) feet
1
171
2
303
3
403
Based on this information, answer the following questions (1 point each).
21.
What is the equation of the
22. What is the initial velocity v0 180 ft/sec?
function that models this data? Use stat
menu, calculate a quadratic regression
What is the initial height h0 7 ft ?
h(t) = -16t2 + 180t + 7
23. What is the height of the projectile at 2.5
seconds? Plug in 2.5 for t or use graph or
table: h(2.5) = 357 ft
24. At what times will the projectile be at
400 feet? Set equal to 400 and solve
(make y2 = 400 and find intersection)
t = 2.96 and 8.285 seconds
25. At what time does the projectile reach its
maximum height? Use maximum feature
in calc menu: t = 5.625 seconds
26. At what time will the projectile reach the
ground? Make y2 = 0 and find
intersection: t = 11.29 seconds
What is its max height? h = 513.25 ft
27. What is a reasonable domain for this
problem?
0 seconds ≤ t ≤ 11.29 seconds
What is a reasonable range for this
problem?
0 ft ≤ h ≤ 513.25 ft
29 – 32 (2 points each)
28. Sketch an accurate labeled and scaled
graph of the problem.
UNIT 1 PRACTICE TEST - QUADRATICS
MATH 2
29. Solve the quadratic equation by
completing the square - show work
x2 – 12x + 7 = 0
x2 – 12x = -7
x2 – 12x + (12/2)2 = -7 + (12/2)2
x2 – 12x + 36 = -7 + 36
(x – 6)2 = 29
x–6=
x=6
x = 11.385 and 0.615
31. For the quadratic function
f(x) = 9x2 – 42x + 49
a. Find the value of the discriminant
b2 – 4ac = (-42)2 – 4(9)(49) = 0
b. Based on your answer to part a,
describe the type of roots the
function has (do not actually solve for
them).
1 – real rational root
NAME_____________________________
DATE______________
30. Solve the quadratic equation by using the
quadratic formula – show work
-2x2 +11x = 3
-2x2 + 11x – 3 = 0
[a = -2, b = 11, c = -3]
 (11)  (11)2  4(2)(3)
x
2(2)
x
 11  97
 x = 0.288 and 5.212
4
32. Solve the following equation by either
factoring, completing the square, or using
the quadratic formula (you choose!)
15x2 = 7x + 2
15x2 – 7x – 2 = 0
Methods will vary:
Solution: x = {-1/5, 2/3}
33. Solve the following system of equations algebraically - write your final answers as (x, y)
coordinates. Show work. (3 points)
Plug values into either equation:
2
2
x + 10x – 3 = 2x – 18
y = 2x – 18 = 2(-3) – 18 = -24
y  x  10x  3

x2 + 8x + 15 = 0
One solution: (-3, -24)
y  2 x  18
(x + 3)(x + 5) = 0
y = 2x – 18 = 2(-5) – 18 = -28
x = -3 x = -5
other solution: (-5, -28)
Solution set: {(-3, -24), (-5, -28)}
34. Solve the following system of equations
35. Graph and shade the solution region of
graphically. Sketch the graph and include
the system of inequalities. (2
2
the points of intersection. (2 points)
points) y  x  2x  8

2

y   x  5 x  1
y  2 x 2  4 x


1
y   x  5
2

Points of
intersection:
(0.816, 4.592)
and
(-3.066, 6.533)