Name: ________________________ Class: ___________________ Date: __________
ID: A
Unit four review
Short Answer
1. Graph the quadratic function y = 3x 2 − 6x + 5 . Use the graph to determine the zeros of the function if they
exist.
2. For what values of k does the equation x 2 + 9x + k = 0 have
a) one real root?
b) two distinct roots?
c) no real roots?
3. Find the value of k that makes the expression x2 + 52x + k a perfect square trinomial.
4. Factor the quadratic 6 (x − 5 ) + 126 (x − 5 ) + 324 completely.
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5. Solve the quadratic function y = 5x 2 + 20x − 6 by completing the square. Round roots to the nearest
hundredth, if necessary.
6. Use the quadratic formula to find the roots of the equation x2 + 4x – 21 = 0. Express your answers as exact
roots.
7. Determine the number of real roots for the equation 3x2 = 8x – 4. Then, find the roots of the equation by
a) using the quadratic formula
b) factoring
8. Find the x-intercepts of the quadratic function y = 3x2 – 10x + 6. Express your answers as exact values.
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Name: ________________________
ID: A
9. Write the equation of this parabola.
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10. Water spraying from a large fountain follows a path modelled by the function h (d ) = − d 2 + 4d for h > 0,
3
where h is the height of the water above the fountain jet and d is the horizontal distance of the water from the
fountain jet, both in metres. How far does the water travel horizontally before splashing back into the
fountain?
Problem
1. The path of a ball can be modelled by the function h (d ) = −2d 2 + 0.5d , where h is the height of the ball and
d is the horizontal distance travelled, both in metres. What total horizontal distance does the ball travel?
2. The path of a parabolic arch is given by h(d) = −0.025d 2 + d , where h is the height of the arch above the
ground, and d is the horizontal width of the arch from the left base, both in metres. How far is the right base
from the left?
3. A baseball batter hits a line drive. The height, h, in metres, of the baseball after t seconds is approximately
modelled by the function h(t) = –5t2 + 45t + 1.
a) Graph the function.
b) State the domain and range of the function.
c) How long does it take for the ball to hit the ground?
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Name: ________________________
ID: A
4. A ball is thrown straight up from a bridge over a river and falls into the water. The height, h, in metres, of the
ball above the water t seconds after being thrown is approximately modelled by the relation
h = −5t 2 + 10t + 35.
a) What is the maximum height of the ball above the water?
b) How long does it take for the ball to reach the maximum height?
c) After how many seconds does the ball hit the water?
d) How high is the bridge above the river?
5. On a forward somersault dive, Nina’s height, h, in metres, above the water is approximately modelled by the
relation h = −5t 2 + 7t + 4, where t is the time in seconds after she leaves the board.
a) Find Nina’s maximum height above the water.
b) How long does it take her to reach the maximum height?
c) How long is it before she enters the water?
d) How high is the board above the water?
6. The height, h, in metres, of an infield fly ball t seconds after being hit is approximately modelled by the
quadratic relation h = 30t − 5t 2 . How long is the ball in the air?
7. The curve of a cable on a suspension bridge can be approximately modelled by the relation
2
h = 0.0025 (d − 100) − 25. The points where the cable attaches to the supporting towers are where h = 0.
How far apart are the supporting towers?
8. Consider the equation 0 = –2(x – 12)2 + 18.
a) List the steps required to find the roots of the quadratic equation.
b) State the roots of the quadratic equation.
9. A golf ball is hit from ground level. Its path is modelled by the relation h = −4.9t 2 + 30.6t , where h is the
ball’s height above the ground, in metres, and t is the time, in seconds. Use the quadratic formula to
determine the time the ball is in the air.
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ID: A
Unit four review
Answer Section
SHORT ANSWER
1. ANS:
There are no zeros.
PTS: 1
DIF: Average
OBJ: Section 4.1
TOP: Graphical Solutions of Quadratic Equations
2. ANS:
a) k = 20.25
b) k < 20.25
c) k > 20.25
NAT: RF 5
KEY: zeros | x-intercepts
PTS: 1
DIF: Difficult
TOP: The Quadratic Formula
3. ANS:
k = 676
OBJ: Section 4.4 NAT: RF 5
KEY: number of roots
PTS: 1
DIF: Easy
TOP: Factoring Quadratic Equations
OBJ: Section 4.2 NAT: RF 5
KEY: perfect square trinomial
1
ID: A
4. ANS:
Let P = (x − 5 ) so that the quadratic becomes 6P 2 + 126P + 324.
Factor the resulting expression:
6P 2 + 126P + 324 = 6(P 2 + 21P + 54)
= 6(P + 3)(P + 18)
= 6[(x − 5) + 3][(x − 5) + 18]
= 6(x − 2)(x + 13)
PTS: 1
DIF: Difficult
OBJ: Section 4.2 NAT: RF 5
TOP: Factoring Quadratic Equations
KEY: polynomials of quadratic form
5. ANS:
To solve the equation, set it equal to 0 and solve for x.
5x 2 + 20x − 6 = 0
x 2 + 4x − 6 / 5 = 0
x 2 + 4x = 6 / 5
x 2 + 4x + (2) = 6 / 5 + (2)
2
2
(x + 2 ) = 26 / 5
2
x + 2 = ± 26 / 5
x = −2 ±
26 / 5
x ≈ 4.28 and x ≈ −0.28
PTS: 1
DIF: Average
OBJ: Section 4.3
TOP: Solving Quadratic Equations by Completing the Square
6. ANS:
x=
x=
=
=
=
−b ±
b 2 − 4ac
2a
−4 ±
4 2 − 4 (1) (−21)
2 (1)
−4 ±
16 + 84
2
−4 ±
100
NAT: RF 5
KEY: completing the square
2
−4 ± 10
2
= 3,−7
PTS: 1
DIF: Easy
TOP: The Quadratic Formula
OBJ: Section 4.4 NAT: RF 5
KEY: quadratic formula
2
ID: A
7. ANS:
Rearrange the equation so all terms are on the same side:
3x 2 − 8x + 4 = 0
Calculate the discriminant b 2 − 4ac:
2
(−8) − 4(3) (4) = 64 − 48
= 16
Since the discriminant is positive (greater than zero), the equation has 2 real roots.
a) x =
−b ±
b 2 − 4ac
2a
=
8 ± 16
2(3)
=
8±4
6
2
3
2
3x − 8x + 4 = 0
= 2 and
b)
(3x − 2) (x − 2 ) = 0
3x − 2 = 0
x−2= 0
3x = 2
x=
x=2
2
3
PTS: 1
DIF: Average
OBJ: Section 4.3 | Section 4.4
NAT: RF 5
TOP: Factoring Quadratic Equations | The Quadratic Formula
KEY: roots of quadratic equation | solve factored trinomial
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ID: A
8. ANS:
x=
=
=
=
=
=
=
−b ±
b 2 − 4ac
2a
10 ±
(−10) − 4 (3) (6)
2 (3)
10 ±
100 − 72
6
10 ±
28
2
6
10 ± 2 7
6
5±
7
3
5+
7
3
and
5−
7
3
PTS: 1
DIF: Average
OBJ: Section 4.4 NAT: RF 5
TOP: The Quadratic Formula
KEY: roots of quadratic equation | quadratic formula
9. ANS:
The x-intercepts are −2 and 3. These correspond to factors of (x + 2 ) and (x − 3 ) . The equation is of the form
y = a(x + 2)(x – 3).
Expand and simplify the right side of the equation:
y = a (x + 2 ) (x − 3 )
= a(x 2 − x − 6)
Substitute the known point on the curve (0.5,−6.25) to determine the value of a:
y = a(x 2 − x − 6)
−6.25 = a[ (0.5) − 0.5 − 6]
2
−6.25 = a(0.25 − 6.5)
−6.25 = a(−6.25)
a=1
The value of a is 1, so the equation is y = x 2 − x − 6 .
PTS: 1
DIF: Easy
OBJ: Section 4.1
TOP: Graphical Solutions of Quadratic Equations
4
NAT: RF 5
KEY: quadratic function | parabola
ID: A
10. ANS:
Factor the equation to determine the zeros:
2
h (d ) = − d 2 + 4d
3
ÊÁ 1
ˆ˜
= 2d ÁÁÁÁ − d + 2 ˜˜˜˜
Ë 3
¯
Set each factor equal to zero and solve for d.
1
2d = 0 and − d + 2 = 0
3
d=0
1
− d = −2
3
d=6
The water spray starts at 0 and ends at 6, so the total horizontal distance is 6 m.
PTS: 1
DIF: Average
TOP: Factoring Quadratic Equations
OBJ: Section 4.2 NAT: RF 5
KEY: solve factored trinomial
PROBLEM
1. ANS:
Graphical solution
Determine the zeros of the function (or roots of the equation) by setting h = 0 and then factoring the equation:
0 = −2d 2 + 0.5d
0 = d (−2d + 0.5)
d = 0 or − 2d + 0.5 = 0
d = 0 or d = 0.25
The ball travels 0.25 m or 25 cm horizontally.
PTS: 1
DIF: Difficult
OBJ: Section 4.1 | Section 4.2
NAT: RF 5
TOP: Graphical Solutions of Quadratic Equations | Factoring Quadratic Equations
KEY: zeros | x-intercepts
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ID: A
2. ANS:
Graphical solution
Determine the zeros of the function (or roots of the equation) by setting h = 0 and then factoring the equation:
0 = −0.025d 2 + d
= d (−0.025d + 1)
−0.025d + 1 = 0
d = 0 or
−0.025d = −1
d = 40
The right base is 40 m from the left base.
PTS: 1
DIF: Difficult
OBJ: Section 4.1 | Section 4.2
NAT: RF 5
TOP: Graphical Solutions of Quadratic Equations | Factoring Quadratic Equations
KEY: x-intercepts | zeros
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ID: A
3. ANS:
a)
b) From the graph, the t-values go from t = 0 to approximately t = 9. Thus, the domain is {t|0 ≤ t ≤ 9,t ∈ R}.
From the graph, the maximum value is approximately 106. Thus, the range is {h|0 ≤ h ≤ 106,h ∈ R}.
c) It takes approximatley 9 s for the ball to hit the ground.
PTS: 1
DIF: Easy
OBJ: Section 4.1
TOP: Graphical Solutions of Quadratic Equations
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NAT: RF 5
KEY: x-intercepts | zeros | domain | range
ID: A
4. ANS:
Graph the relation to visualize the situation.
From the graph:
a) The maximum height of the ball is 40 m.
b) It takes 1 s to reach the maximum height.
c) The t-intercept is approximately 3.8. The ball hits the ground after about 3.8 s.
From the equation:
d) When t = 0, h = 35. The bridge is 35 m above the river.
PTS: 1
DIF: Average
OBJ: Section 4.1
TOP: Graphical Solutions of Quadratic Equations
KEY: maximum | x-intercepts | parabolic motion
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NAT: RF 5
ID: A
5. ANS:
Graph the relation to visualize the situation.
From the graph:
a) The maximum height is about 6.5 m.
b) It takes about 0.7 s to reach the maximum height.
c) The t-intercept is about 1.8. It takes Nina about 1.8 s to enter the water.
d) Substituting t = 0 into the equation, or reading from the graph at t = 0, h = 4. So, the board is 4 m above
the water.
PTS: 1
DIF: Difficult
OBJ: Section 4.1
TOP: Graphical Solutions of Quadratic Equations
KEY: maximum | x-intercepts | parabolic motion
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NAT: RF 5
ID: A
6. ANS:
The ball is in the air until h = 0.
Factor the trinomial to determine when this occurs.
0 = 30t − 5t 2
0 = 5t (6 − t )
t = 0 or t = 6
The zeros occur at t = 0 and t = 6.
The ball is in the air for 6 s.
PTS: 1
DIF: Easy
TOP: Factoring Quadratic Equations
7. ANS:
2
0 = 0.0025 (d − 100) − 25
25 = 0.0025 (d − 100)
10 000 = (d − 100)
OBJ: Section 4.2 NAT: RF 5
KEY: factor quadratic
2
2
±100 = d − 100
d = 100 ± 100
d = 0 or d = 200
The supporting towers are 200 m apart.
PTS: 1
DIF: Average
OBJ: Section 4.3 NAT: RF 5
TOP: Solving Quadratic Equations by Completing the Square KEY: completing the square
8. ANS:
a) Simplify the equation by dividing both sides by –2, factor the difference of squares, simplify, read the
roots from the factored equation.
b) 9, 15
0 = −2(x − 12) 2 + 18
0 = (x − 12) 2 − 9
0 = (x − 12 − 3)(x − 12 + 3)
0 = (x − 15)(x − 9)
x = 15 or x = 9
PTS: 1
DIF: Average
OBJ: Section 4.3
TOP: Solving Quadratic Equations by Completing the Square
10
NAT: RF 5
KEY: completing the square
ID: A
9. ANS:
t=
−30.6 ±
(30.6) − 4 (−4.9) (0)
2 (−4.9)
2
−30.6 ± 30.6
−9.8
t = 0 and t ≈ 6.2
The ball is in the air for approximately 6.2 s.
=
PTS: 1
DIF: Easy
TOP: The Quadratic Formula
OBJ: Section 4.4 NAT: RF 5
KEY: quadratic formula
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