Download ExamView - Final Review - Unit 7A.tst

Matching
Final Review - Unit 7A
Match each function to the corresponding graphical representation below.
A
D
Completion
Complete each statement.
1. The y-intercept of the graph of f (x ) = |−5x + 9| is ______________________________.
|
|
2
2. The range of the function f (x ) = | 6 (x − 7 ) + 7 | is ______________________________.
3. The invariant points for the function f (x ) =
1
are_____________________________ and
x+2
_____________________________.
B
C
4. f (x ) = |2x + 8|
E
5. g (x ) = || x 2 − 5x + 3 ||
Match each system of equations to the corresponding graphical representation below.
D
A
2
|
|
6. h (x ) = | (x − 2 ) − 5 |
7. k (x ) =
1
2x + 8
8. j(x) =
1
x 2 − 5x + 3
B
E
C
9. y = −1.5x 2 + 1.5x + 3
y = 1.5(x + 2.3 ) + 2.5
2
10. y = 1.3x + 3
y = 1.5(x + 2.3 ) + 2.5
2
11. y = 1.5x 2 + 1.5x + 3
y = −1.5 (x − 2.3 ) − 2.5
2
12. y = −1.3x + 3
y = 1.5(x + 2.3 ) + 2.5
2
13. y = −1.5x 2 + 1.5x + 3
y = 1.5(x − 2.3 ) − 2.5
2
Short Answer
14. Evaluate each absolute value expression.
a) 6 + |5 − 11|
b) −2 − |7| + |3 − 2|
24
c) |
− | 12 ÷ (−2) ||
d) |2| × ÁÊË − |−3| ˜ˆ¯ × (−2)
15. Consider the function f (x ) = || 2x 2 − 16x + 29 || .
|
|
2
a) Express the function in vertex form, y = || a ÊÁË x − p ˆ˜¯ + q || .
b) Sketch the graph of the function. Label the vertex and x-intercepts.
c) What are the domain and range?
|1
|
16. Solve the absolute value equation | x + 1 | = x + 1 graphically.
|2
|
| 2
|
x
− 1 | = x + 3 algebraically.
17. Determine the solution to the equation |
|| 2
||
18. Graph each reciprocal function. For each graph, state the non-permissible values and the equation of the vertical
asymptote(s).
1
a) y =
3x − 2
1
b) y = 2
x − 16
19. Solve |x + 2 | ≥
1
.
x+2
20. Solve the system graphically.
y = 3x − 7
y = x2 − x − 3
21. Determine the number of points of intersection for the line y = 3x + 5 and the curve y = 2x 2 + 4x − 1 .
22. Determine if each system has one, two, or no solutions.
a) y = 2x 2 − 2x + 1 and y = 3x − 5
b) y = x 2 + 3x − 16 and y = −x 2 − 8x − 18
23. Solve the system of equations by elimination.
y = 2x 2 − 2x − 3 and y = −x 2 − 2x − 3
24. Solve the system of equations using substitution. State your answers to two decimal places.
y = −3x 2 − 3x + 2 and y = −6x 2 + 4x + 7
Problem
25. The temperature of an enclosure for a pet snake should be about 27°C, give or take 3°C.
a) What absolute value expression can be used to represent this situation?
b) Use the expression to determine the temperature range the enclosure should have.
26. Green plants live in the ocean at maximum and minimum depths described by the absolute value equation
|15 − d | − 15 = 0, where d is the depth, in metres. Graph the equation on a number line to determine the maximum
and minimum depths at which green plants live in the ocean.
34. A rectangle has an area of 35 cm2 with a side length that is 2 cm longer than its width.
a) Identify the system of equations that represents this situation.
b) Graph the system and determine the dimensions of the rectangle.
27. What is the area of the triangle formed between the graph of f (x ) = |2x − 4| and the line y = 6 ? Include a graph of
the situation.
28. Is the statement || f (x ) || = f( |x |) always true, sometimes true, or never true? Use examples to explain your answer.
29. Solve the equation |x − 1 | + |x + 5 | = 6 algebraically and then graph the solution to check.
30. For which values of k does the equation || x 2 − 9 || = −x + k have no solution?
31. What is the equation of the function in the form f (x ) =
1
for the function with a vertical asymptote at x = 1?
x−c
Sketch the function.
32. Find the values of x in the numbers x, x + 1, and x + 2 such that the reciprocal of the smallest number equals the
sum of the reciprocals of the other two.
33. Graph the reciprocal function f (x ) =
1
and determine
x2 + x − 6
a) the equations of any asymptotes
b) the domain and range
35. Determine the value of k in y = kx 2 − 5x + 2 that will result in the intersection of the line y = −3x + 4 with the
quadratic at
a) two points
b) one points
c) no point
36. The sum of the squares of two numbers is 74 and the difference of their squares is 24.
a) What system of equations models this situation?
b) What are the two numbers?
37. Solve the system algebraically.
y = 4x 2 + 13
y + 7 = 4x 2
38. The sum of the reciprocals of two numbers is 4. Four times the product of their reciprocals is 15. What are the two
numbers?
ID: A
ID: A
15. a) f (x ) = || 2x 2 − 16x + 29 ||
Final Review - Unit 7A
Answer Section
= || 2(x 2 − 8x) + 29||
= || 2(x 2 − 8x + 16 − 16) + 29 ||
COMPLETION
2
|
|
= | 2 (x − 4 ) − 32 + 29 |
1. 9
ÏÔ
¸Ô
2. ÔÌ y| y ≥ 7, y ∈ R Ô˝
Ó
˛
3. ÁÊË −1,1 ˜ˆ¯ and ÁÊË −3,−1 ˜ˆ¯
2
|
|
= | 2 (x − 4 ) − 3 |
b)
MATCHING
4.
5.
6.
7.
8.
B
A
E
C
D
9.
10.
11.
12.
13.
B
A
E
C
D
ÔÏ
Ô¸
c) The domain is all real numbers or ÌÔ x| x ∈ R ˝Ô .
Ó
˛
ÏÔ
¸Ô
The range is all non-negative values of y, or ÔÌ y| y ≥ 0, y ∈ R Ô˝ .
Ó
˛
SHORT ANSWER
14. a) 6 + |5 − 11| = 6 + |−6|
= 6+6
= 12
b) −2 −|7|+|3 − 2| = −2 − 7 + |1|
= −9 + 1
= −8
24
24
=
c) |
− | 12 ÷ (−2) || − |−6|
=
24
−6
= −4
d) |2| × ÊÁË − |−3| ˆ˜¯ × (−2) = 2 (−3) (−2)
= 12
ID: A
|1
|
16. Graph y = | x + 1 | and y = x + 1 and find their point(s) of intersection.
|2
|
ID: A
x2
−1≥ 0
2
| 2
|
| x − 1| = x + 3
|| 2
||
17. Case 1:
x2
−1= x+3
2
x 2 − 2 = 2x + 6
x − 2x − 8 = 0
2
(x − 4 ) (x + 2 ) = 0
x = 4 or x = −2
Check x = 4:
x2
L.S. =
−1
2
=
42
−1
2
R.S. = x + 3
= 4+3
=7
= 8−1
|
|1
The point of intersection of the two graphs is (0,1). x = 0 is the solution to | x + 1 | = x + 1.
|
|2
=7
L.S. = R.S.
Check x = −2:
x2
L.S. =
−1
2
=
(−2) 2
−1
2
R.S. = x + 3
= −2 + 3
=1
= 2−1
=1
L.S. = R.S.
x2
−1< 0
Case 2:
2
| 2
|
| x − 1| = x + 3
|| 2
||
x2
− 1 = −x − 3
2
x 2 − 2 = −2x − 6
x + 2x + 4 = 0
2
There is no real solution to this quadratic because the discriminant, 2 2 − 4 (1) (4) = −12, is negative.
Therefore, the solution is x = 4 or x = −2.
ID: A
18. a)
ID: A
19. Graph the functions y = |x + 2 | and y =
From the graph, | x + 2| ≥
The non-permissible value occurs when 3x − 2 = 0 or when x =
The equation of the vertical asymptote is x =
2
.
3
2
.
3
b)
20.
The solution is (2,−1).
The non-permissible values occur when x 2 − 16 = 0 or when x = ±4.
The equations of the vertical asymptotes are x = 4 and x = −4.
1
.
x+2
1
when x < −2 and when x ≥ −1.
x+2
ID: A
21. Equate the expressions and simplify.
2x 2 + 4x − 1 = 3x + 5
2x 2 + x − 6 = 0
Use the discriminant b 2 − 4ac to determine the number of solutions to this equation.
Substitute a = 2, b = 1, and c = −6.
b 2 − 4ac = (1) 2 − 4(2)(−6)
= 1 + 48
= 49
Since b 2 − 4ac > 0, there are two solutions to the equation.
The line and the curve have two points of intersection.
22. a) Equate the expressions and simplify.
2x 2 − 2x + 1 = 3x − 5
2x 2 − 5x + 6 = 0
Use the discriminant b 2 − 4ac to determine the number of solutions to this equation.
Substitute a = 2, b = −5, and c = 6.
b 2 − 4ac = (−5) 2 − 4(2)(6)
= 25 − 48
= −23
Since b 2 − 4ac < 0, there are no solutions to the equation.
The system has no solutions.
b) Equate the expressions and simplify.
x 2 + 3x − 16 = −x 2 − 8x − 18
2x 2 + 11x + 2 = 0
Use the discriminant b 2 − 4ac to determine the number of solutions to this equation.
Substitute a = 2, b = 11, and c = 2.
b 2 − 4ac = (11) 2 − 4(2)(2)
= 121 − 16
= 105
Since b 2 − 4ac > 0, there are two solutions to the equation.
The system has two solutions.
23. Subtract the equations:
y = 2x 2 − 2x − 3
y = −x 2 − 2x − 3
0 = 3x 2
Solve for x:
0 = 3x 2
x=0
Substitute x = 0 into either equation and solve for y:
2
y = − (0) − 2 (0) − 3
= −3
The single solution is (0, −3).
ID: A
24. Substitute y = −6x 2 + 4x + 7 into the first equation:
−6x 2 + 4x + 7 = −3x 2 − 3x + 2
0 = 3x 2 − 7x − 5
Solve for x using the quadratic formula
x=
7±
(−7) − 4 (3) (−5)
2 (3)
2
7±
109
6
x ≈ 2.91 and x ≈ −0.57
Substitute these values into y = −3x 2 − 3x + 2 :
=
y ≈ −3 (2.91) − 3 (2.91) + 2
2
and
y ≈ −3 (−0.57) − 3 (−0.57) + 2
2
≈ −32.13
≈ 2.74
The approximate solutions are (2.91, −32.13) and (−0.57, 2.74).
PROBLEM
25. a) If t is the temperature of the enclosure, then |t − 27| = 3 represents the temperature range.
b) t − 27 = 3 or t − 27 = −3
t = 30
t = 24
The temperature of the enclosure should be between 24°C and 30°C.
26.
The maximum depth is 30 m and the minimum depth is 0 m.
ID: A
ID: A
28. The statement is sometimes true. Using f (x ) = −2x + 1 , the red graph is || f (x ) || , while the blue graph is f( |x |) :
Using f (x ) = 2x as the starting function, the graph of || f (x ) || is the same as the graph of f( |x |) :
27.
From the graph, the base of the triangle is 5 − (−1) = 6 units and the height of the triangle is 6 units. The area of
the triangle is
1
A = bh
2
=
1
(6) (6)
2
= 18
The triangle has an area of 18 square units.
Whenever vertical or horizontal translations are involved, || f (x ) || ≠ f( |x |).
ID: A
29. Case 1 (x < −5):
If x < −5, then both expressions within the absolute value bars are negative. Thus,
−(x − 1) + [−(x + 5)] = 6
ID: A
30. Draw the graphs of || x 2 − 9 || = y and y = −x + k for several values of k on the same grid.
Determine which values of k do not permit the intersection of the two graphs
−x + 1 − x − 5 = 6
−2x = 10
x = −5
This solution is extraneous, because in this case x < −5.
Case 2 (−5 ≤ x < 1):
If −5 ≤ x < 1, the first expression is negative and the second expression is greater than or equal to 0. Thus,
−(x − 1) + x + 5 = 6
−x + 1 + x + 5 = 6
6=6
This is true for all values of x in the interval.
Case 3 (x ≥ 1):
If x ≥ 1, then both expressions are greater than or equal to 0. Thus,
x−1+x +5 = 6
2x = 2
x=1
Thus, the equation is true for −5 ≤ x < 1 and x = 1. These solutions are all included in the interval −5 ≤ x ≤ 1.
Thus, the solution to the equation is −5 ≤ x ≤ 1.
Any value of k < −3 will result in an equation with no solution.
ID: A
31. A vertical asymptote of x = 1 corresponds to c = 1 or f (x ) =
1
.
x−1
ID: A
32. The situation can be modelled by the equation
1
1
1
=
+
x x+1 x+2
(x + 1)(x + 2) = x(x + 2) + x(x + 1)
x 2 + 3x + 2 = x 2 + 2x + x 2 + x
x2 = 2
x=± 2
1
1
1
=
+
.
x x+1 x+2
ID: A
ID: A
34. a) Set up and solve a system of equations. Let l and w represent the length and width, respectively, of the
rectangle.
lw = 35
l = w+2
Graph the equations.
33.
a) f(x) =
1
x2 + x − 6
1
(x + 3)(x − 2)
There are vertical asymptotes at x = 2 and x = −3 and a horizontal asymptote at y = 0.
ÔÏ
Ô¸
b) Domain ÌÔ x| x ≠ 2,−3, x ∈ R ˝Ô
Ó
˛
¸Ô
ÏÔ
Range ÔÌ y| y ≠ 0, y ∈ R Ô˝
˛
Ó
=
From the graph, the solution is (5, 7). Disregard the second solution because it involves negative distances.
The rectangle is 5 cm wide and 7 cm long.
ID: A
35. −3x + 4 = kx 2 − 5x + 2
0 = kx 2 − 2x − 2
The discriminant is
(−2) 2 − 4(k)(−2) = 4 + 8k
a) For two points of intersection, the discriminant must be greater than zero:
4 + 8k > 0
8k > −4
1
2
b) For one point of intersection, the discriminant must be equal to zero:
4 + 8k = 0
k >−
8k = −4
ID: A
38. Set up a system of equations using the information in the problem. Let x and y represent the two numbers.
1 1
+ =4
x y
ÁÊ 1 ˜ˆ ÁÊ 1 ˜ˆ
4 ÁÁÁ ˜˜˜˜ ÁÁÁÁ ˜˜˜˜ = 15
ÁË x ¯ Ë y ¯
Re-write the equations as follows:
ÁÊ 1 ˜ˆ ÁÊ 1 ˜ˆ
1 1
+ =4
4 ÁÁÁÁ ˜˜˜˜ ÁÁÁÁ ˜˜˜˜ = 15
and
x y
Ë x ¯Ë y ¯
y
x
+
=4
xy xy
y + x = 4xy
(1)
Substitute (2) into (1).
1
2
c) For no points of intersection, the discriminant must be less than zero:
4 + 8k < 0
k =−
ÁÊ 4 ˜ˆ˜
4
˜˜
+ x = 4x ÁÁÁÁ
˜
15x
Ë 15x ¯
4
16
+x =
15x
15
8k < −4
k <−
1
2
36. a) x 2 + y 2 = 74
x 2 − y 2 = 24
b) Add the two equations:
x 2 + y 2 = 74
x − y = 24
2
2
2x 2 = 98
x 2 = 49
x = ±7
Substitute these values into x 2 + y 2 = 74 and solve for y.
(±7) + y 2 = 74
2
y 2 = 25
y = ±5
The two numbers are 7 and 5, 7 and −5, −7 and 5, or −7 and −5.
37. 4x 2 − 7 = 4x 2 + 13
0 = 20
Since this is impossible, there is no solution to this system of equations.
4
=y
15x
4 + 15x 2 = 16x
15x 2 − 16x + 4 = 0
(3x − 2) (5x − 2) = 0
2
2
and x =
3
5
x=
The two numbers are
2
2
and .
3
5
(2)