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Deck 4: Polynomial and Rational Functions Applications to Optimization
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Question
Find all fixed points of the function. 
A) x = 17
B) x = 0, x = 170
C) x = 10
D) x = 17, x = 10
E) no fixed points
A) x = 17
B) x = 0, x = 170
C) x = 10
D) x = 17, x = 10
E) no fixed points
Question
Suppose that the revenue generated by selling x units of a certain commodity is given by 
. Assume that R is in dollars. What is the maximum revenue possible in this situation?
A) $450,000
B) $440,000
C) $430,000
D) $900,000
E) $480,000
. Assume that R is in dollars. What is the maximum revenue possible in this situation?A) $450,000
B) $440,000
C) $430,000
D) $900,000
E) $480,000
Question
Sketch the graph of the function and specify all x- and y-intercepts. y = - 3x 4 + 5
A)
x-intercepts: 5
Y-intercept: - 50,625
B)
x-intercepts: -5
Y-intercept: - 50,625
C)
x-intercepts: 
Y-intercept: - 5
D)
x-intercepts: 
Y-intercept: 5
E)
x-intercept: 0
Y-intercept: 0
A)
x-intercepts: 5Y-intercept: - 50,625
B)
x-intercepts: -5Y-intercept: - 50,625
C)
x-intercepts: Y-intercept: - 5
D)
x-intercepts: Y-intercept: 5
E)
x-intercept: 0Y-intercept: 0
Question
Graph the quadratic function. Specify the vertex, axis of symmetry, maximum or minimum value, and intercepts. 
A) vertex: (- 2, 0); axis of symmetry: t = - 2; maximum value: 0; t-intercept: - 2; s-intercept: -1.
B) vertex: (- 2, 0); axis of symmetry: t = - 2; minimum value: 0; t-intercept: - 2; s-intercept: 1.
C) vertex: (2, 0); axis of symmetry: t = 2; maximum value: 0; t-intercept: 2; s-intercept: -1.
D) vertex: (0,1); axis of symmetry: t = 0; maximum value: 1; t-intercept:
; s-intercept: 1. 
E) vertex: (0, 2);axis of symmetry: t = 0; maximum value: 2; t-intercept:
;s-intercept: 2. 
A) vertex: (- 2, 0); axis of symmetry: t = - 2; maximum value: 0; t-intercept: - 2; s-intercept: -1.
B) vertex: (- 2, 0); axis of symmetry: t = - 2; minimum value: 0; t-intercept: - 2; s-intercept: 1.
C) vertex: (2, 0); axis of symmetry: t = 2; maximum value: 0; t-intercept: 2; s-intercept: -1.
D) vertex: (0,1); axis of symmetry: t = 0; maximum value: 1; t-intercept:
; s-intercept: 1. E) vertex: (0, 2);axis of symmetry: t = 0; maximum value: 2; t-intercept:
;s-intercept: 2. Question
Sketch the graph of the function and specify all x- and y-intercepts. y =- ( x - 3) 3 - 1
A)
x-intercept: -1
Y-intercept: -1
B)
x-intercept: -2
Y-intercept: 26
C)
x-intercept: 2
Y-intercept: 26
D)
x-intercept: 2
Y-intercept: - 26
E)
x-intercept: 3
Y-intercept: 27
A)
x-intercept: -1Y-intercept: -1
B)
x-intercept: -2Y-intercept: 26
C)
x-intercept: 2Y-intercept: 26
D)
x-intercept: 2Y-intercept: - 26
E)
x-intercept: 3Y-intercept: 27
Question
A triangle is inscribed in a semicircle of diameter 6R. Show that the smallest possible value for the area of the shaded region is 
. 
Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle.
A) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to
.
The square of the area of the triangle is equal to
, and the substitution 
will transform this expression into the quadratic function 
Since we want to find the maximum value of t, we will substitute the value
into the equation. Solving for t gives us the following minimum area of the shaded region: 
.
B) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to
.
The square of the area of the triangle is equal to
The substitution 
will transform this expression into the quadratic function 
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value
into the equation 
gives us 
and consequently 
(The negative root can be rejected since the side of a triangle can't be negative). With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to 
Substituting the value
in the equation (2) gives us that the minimum value of the shaded region is equal to 
.
C) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to
.
The substitution
will transform this into the quadratic function 
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value
into the equation 
gives us 
and consequently 
. With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to 
Substituting the value
into the equation (2), we find that the minimum value of the shaded region is equal to 
.
D) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle.
The square of the area of the triangle is equal to
, which is a quadratic function. The graph of this function will be a parabola opening downward, so we can write the maximum value of this function as: 
We can then write
as 
and calculate the minimum area of the shaded region. Substituting this value into the area equation, we find its minimum area: 
.
E) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to
.
The substitution
will transform this into the quadratic function 
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value
into the equation 
gives us 
and consequently 
. With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to 
Substituting the value
into the equation (2), we find that the minimum value of the shaded region is equal to 
.
. Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle.A) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to
.The square of the area of the triangle is equal to
, and the substitution will transform this expression into the quadratic function Since we want to find the maximum value of t, we will substitute the value
into the equation. Solving for t gives us the following minimum area of the shaded region: .B) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to
.The square of the area of the triangle is equal to
The substitution will transform this expression into the quadratic function Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value
into the equation gives us and consequently (The negative root can be rejected since the side of a triangle can't be negative). With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to Substituting the value
in the equation (2) gives us that the minimum value of the shaded region is equal to .C) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to
.The substitution
will transform this into the quadratic function Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value
into the equation gives us and consequently . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to Substituting the value
into the equation (2), we find that the minimum value of the shaded region is equal to .D) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle.
The square of the area of the triangle is equal to
, which is a quadratic function. The graph of this function will be a parabola opening downward, so we can write the maximum value of this function as: We can then write
as and calculate the minimum area of the shaded region. Substituting this value into the area equation, we find its minimum area: .E) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to
.The substitution
will transform this into the quadratic function Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value
into the equation gives us and consequently . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to Substituting the value
into the equation (2), we find that the minimum value of the shaded region is equal to . Question
Plot the following points: (6, 2), (7, 5), (8, 8), (9, 9). In your scatter diagram, sketch a line that best seems to fit the data. Estimate the slope and the y-intercept of the line.
A) Estimated slope: 2; Estimated y-intercept: -12.5
B) Estimated slope: 2.5; Estimated y-intercept: -13
C) Estimated slope: 2.9; Estimated y-intercept: -15.3
D) Estimated slope: 1.2; Estimated y-intercept: -1.5
E) Estimated slope: 2.5; Estimated y-intercept: -16
A) Estimated slope: 2; Estimated y-intercept: -12.5
B) Estimated slope: 2.5; Estimated y-intercept: -13
C) Estimated slope: 2.9; Estimated y-intercept: -15.3
D) Estimated slope: 1.2; Estimated y-intercept: -1.5
E) Estimated slope: 2.5; Estimated y-intercept: -16
Question
Sketch the graph of the function and specify all x- and y-intercepts. y =(x - 2) 2 + 4
A)
There is no x-intercept. y-intercept: 8
B)
There is no x-intercept. y-intercept: 8
C)
There is no x-intercept. y-intercept: 7
D)
x-intercept: 0 y-intercept: 4
E)
There is no x-intercept. y-intercept: 7
A)
There is no x-intercept. y-intercept: 8B)
There is no x-intercept. y-intercept: 8C)
There is no x-intercept. y-intercept: 7D)
x-intercept: 0 y-intercept: 4E)
There is no x-intercept. y-intercept: 7 Question
Question
Find the linear function satisfying the given conditions. 
and 
A)
B)
C)
D)
E)
and A)
B)
C)
D)
E)
Question
For the following figure, express the length AB as a function of x. (Hint: Note the similar triangles.) 
A)
B)
C)
D)
E)
A)
B)
C)
D)
E)
Question
Find all fixed points of the function. 
A) x = 1
B) x = - 1
C) x = 4
D) x = 3
E) no fixed points
A) x = 1
B) x = - 1
C) x = 4
D) x = 3
E) no fixed points
Question
A factory owner buys a new machine for $25,000. After eight years, the machine has a salvage value of $1,000. Find a formula for the value of the machine after t years, where 
A) V (t) = - 3,000x + 25,000
B) V (t) = - 3,000x + 1,000
C) V (t) = 3,000x + 25,000
D) V (t) = 3,000x - 25,000
E) V (t) = - 3,000x - 25,000
A) V (t) = - 3,000x + 25,000
B) V (t) = - 3,000x + 1,000
C) V (t) = 3,000x + 25,000
D) V (t) = 3,000x - 25,000
E) V (t) = - 3,000x - 25,000
Question
Sketch the graph of the function and specify all x- and y-intercepts. y = (x - 2)(x - 1)(x + 1)
A)
x-intercepts: - 2, 1, 2
Y-intercept: -4
B)
x-intercepts: - 1, 1, 2
Y-intercept: 2
C)
x-intercepts: - 1, 1, 2
Y-intercept: -2
D)
x-intercepts: - 2, 1, 2
Y-intercept: 4
E)
x-intercepts: - 1, 1
Y-intercepts: - 2.25
A)
x-intercepts: - 2, 1, 2Y-intercept: -4
B)
x-intercepts: - 1, 1, 2Y-intercept: 2
C)
x-intercepts: - 1, 1, 2Y-intercept: -2
D)
x-intercepts: - 2, 1, 2Y-intercept: 4
E)
x-intercepts: - 1, 1Y-intercepts: - 2.25
Question
Two points A and B move along the x-axis. After t sec, their positions are given by the equations 
At what time t do A and B have the same x-coordinate?
A) 7 sec
B) 6 sec
C) 5 sec
At what time t do A and B have the same x-coordinate?A) 7 sec
B) 6 sec
C) 5 sec
Question
Find all fixed points of the function. 
A) x = - 1, x = - 9
B) x = - 1, x = 9
C) x = - 9, x = 1
D) x = 9
E) no fixed points
A) x = - 1, x = - 9
B) x = - 1, x = 9
C) x = - 9, x = 1
D) x = 9
E) no fixed points
Question
Determine the input that produces the largest or smallest output (whichever is appropriate). State whether the output is largest or smallest. s = - 24t 2 + 147t + 15
A)
; largest
B)
; largest
C)
; smallest
D)
; largest
E)
; smallest
A)
; largestB)
; largestC)
; smallestD)
; largestE)
; smallest Question
A piece of wire 
inches long is bent into a circle. Express the area of the circle as a function of y.
A)
B)
C)
D)
E)
inches long is bent into a circle. Express the area of the circle as a function of y.A)
B)
C)
D)
E)
Question
Question
Among all rectangles having a perimeter of 24m, find the dimensions of the one with the largest area.
A)
B)
C)
D)
E)
A)
B)
C)
D)
E)
Question
Sketch the graph of the rational function. Specify the intercepts and the asymptotes. 
A) no x-intercepts; y-intercept: 1; no vertical asymptotes; horizontal asymptote: y = 0;
B) no x-intercepts; y-intercept: 1; vertical asymptote: x = 1; horizontal asymptote: y = 0;
C) no x-intercepts; y-intercept: 1; vertical asymptote: x = - 1; horizontal asymptote: y = 0;
D) no x-intercepts; y-intercept: -1; vertical asymptote: x = 1; horizontal asymptote: y = 0;
E) no x-intercepts; y-intercept: -1; no vertical asymptotes; horizontal asymptote: y = 0;
A) no x-intercepts; y-intercept: 1; no vertical asymptotes; horizontal asymptote: y = 0;
B) no x-intercepts; y-intercept: 1; vertical asymptote: x = 1; horizontal asymptote: y = 0;
C) no x-intercepts; y-intercept: 1; vertical asymptote: x = - 1; horizontal asymptote: y = 0;
D) no x-intercepts; y-intercept: -1; vertical asymptote: x = 1; horizontal asymptote: y = 0;
E) no x-intercepts; y-intercept: -1; no vertical asymptotes; horizontal asymptote: y = 0;
Question
Sketch the graph of the function and specify all x- and y-intercepts.y = x 3 - 9x
A)
x-intercept: 0
Y-intercept: 0
B)
x-intercept: 0
Y-intercept: 0
C)
x-intercepts: -3, 0, 3
Y-intercept: 0
D)
x-intercepts: -3, 0, 3
Y-intercept: 0
E)
x-intercept: 0
Y-intercept: 0
A)
x-intercept: 0Y-intercept: 0
B)
x-intercept: 0Y-intercept: 0
C)
x-intercepts: -3, 0, 3Y-intercept: 0
D)
x-intercepts: -3, 0, 3Y-intercept: 0
E)
x-intercept: 0Y-intercept: 0
Question
Sketch the graph of the rational function. Specify the intercepts and the asymptotes. 
A) x-intercept: 3; y-intercept: - 1; vertical asymptote: x = - 3; horizontal asymptote: y = 1;
B) x-intercept: 0; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = - 1;
C) x-intercept: 3; y-intercept: 1; vertical asymptote: x = - 3; horizontal asymptote: y = - 1;
D) x-intercept: 3; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = - 1;
E) x-intercept: - 3; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = 1;
A) x-intercept: 3; y-intercept: - 1; vertical asymptote: x = - 3; horizontal asymptote: y = 1;
B) x-intercept: 0; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = - 1;
C) x-intercept: 3; y-intercept: 1; vertical asymptote: x = - 3; horizontal asymptote: y = - 1;
D) x-intercept: 3; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = - 1;
E) x-intercept: - 3; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = 1;
Question
Sketch the graph of the function and specify all x- and y-intercepts. y = x 3 (x + 1)
A)
x-intercepts: 0, 1
Y-intercept: 0
B)
x-intercepts: 0, - 1
Y-intercept: 0
C)
x-intercepts: 0, 1
Y-intercept: 0
D)
x-intercepts: 0, 1
Y-intercept: 0
E)
x-intercepts: 0, - 1
Y-intercept: 0
A)
x-intercepts: 0, 1Y-intercept: 0
B)
x-intercepts: 0, - 1Y-intercept: 0
C)
x-intercepts: 0, 1Y-intercept: 0
D)
x-intercepts: 0, 1Y-intercept: 0
E)
x-intercepts: 0, - 1Y-intercept: 0
Question
Sketch the graph of the rational function. Specify the intercepts and the asymptotes. 
A) no x-intercepts; y-intercept:
; vertical asymptote: x = - 4; horizontal asymptote: y = 0; 
B) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; 
C) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; 
D) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; 
E) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; 
A) no x-intercepts; y-intercept:
; vertical asymptote: x = - 4; horizontal asymptote: y = 0; B) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; C) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; D) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; E) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; 
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Deck 4: Polynomial and Rational Functions Applications to Optimization
1
Find all fixed points of the function.
A) x = 17
B) x = 0, x = 170
C) x = 10
D) x = 17, x = 10
E) no fixed points
A) x = 17
B) x = 0, x = 170
C) x = 10
D) x = 17, x = 10
E) no fixed points
x = 17
2
Suppose that the revenue generated by selling x units of a certain commodity is given by . Assume that R is in dollars. What is the maximum revenue possible in this situation?
A) $450,000
B) $440,000
C) $430,000
D) $900,000
E) $480,000
. Assume that R is in dollars. What is the maximum revenue possible in this situation?A) $450,000
B) $440,000
C) $430,000
D) $900,000
E) $480,000
$450,000
3
Sketch the graph of the function and specify all x- and y-intercepts. y = - 3x 4 + 5
A) x-intercepts: 5
Y-intercept: - 50,625
B) x-intercepts: -5
Y-intercept: - 50,625
C) x-intercepts:
Y-intercept: - 5
D) x-intercepts:
Y-intercept: 5
E) x-intercept: 0
Y-intercept: 0
A)
x-intercepts: 5Y-intercept: - 50,625
B)
x-intercepts: -5Y-intercept: - 50,625
C)
x-intercepts: Y-intercept: - 5
D)
x-intercepts: Y-intercept: 5
E)
x-intercept: 0Y-intercept: 0
x-intercepts: Y-intercept: 5
4
Graph the quadratic function. Specify the vertex, axis of symmetry, maximum or minimum value, and intercepts.
A) vertex: (- 2, 0); axis of symmetry: t = - 2; maximum value: 0; t-intercept: - 2; s-intercept: -1.
B) vertex: (- 2, 0); axis of symmetry: t = - 2; minimum value: 0; t-intercept: - 2; s-intercept: 1.
C) vertex: (2, 0); axis of symmetry: t = 2; maximum value: 0; t-intercept: 2; s-intercept: -1.
D) vertex: (0,1); axis of symmetry: t = 0; maximum value: 1; t-intercept: ; s-intercept: 1.
E) vertex: (0, 2);axis of symmetry: t = 0; maximum value: 2; t-intercept: ;s-intercept: 2.
A) vertex: (- 2, 0); axis of symmetry: t = - 2; maximum value: 0; t-intercept: - 2; s-intercept: -1.
B) vertex: (- 2, 0); axis of symmetry: t = - 2; minimum value: 0; t-intercept: - 2; s-intercept: 1.
C) vertex: (2, 0); axis of symmetry: t = 2; maximum value: 0; t-intercept: 2; s-intercept: -1.
D) vertex: (0,1); axis of symmetry: t = 0; maximum value: 1; t-intercept:
; s-intercept: 1. E) vertex: (0, 2);axis of symmetry: t = 0; maximum value: 2; t-intercept:
;s-intercept: 2. Unlock Deck
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5
Sketch the graph of the function and specify all x- and y-intercepts. y =- ( x - 3) 3 - 1
A) x-intercept: -1
Y-intercept: -1
B) x-intercept: -2
Y-intercept: 26
C) x-intercept: 2
Y-intercept: 26
D) x-intercept: 2
Y-intercept: - 26
E) x-intercept: 3
Y-intercept: 27
A)
x-intercept: -1Y-intercept: -1
B)
x-intercept: -2Y-intercept: 26
C)
x-intercept: 2Y-intercept: 26
D)
x-intercept: 2Y-intercept: - 26
E)
x-intercept: 3Y-intercept: 27
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6
A triangle is inscribed in a semicircle of diameter 6R. Show that the smallest possible value for the area of the shaded region is . Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle.
A) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to .
The square of the area of the triangle is equal to , and the substitution will transform this expression into the quadratic function
Since we want to find the maximum value of t, we will substitute the value into the equation. Solving for t gives us the following minimum area of the shaded region: .
B) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to .
The square of the area of the triangle is equal to The substitution will transform this expression into the quadratic function
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value into the equation gives us and consequently (The negative root can be rejected since the side of a triangle can't be negative). With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to
Substituting the value in the equation (2) gives us that the minimum value of the shaded region is equal to .
C) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to .
The substitution will transform this into the quadratic function
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value into the equation gives us and consequently . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to
Substituting the value into the equation (2), we find that the minimum value of the shaded region is equal to .
D) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle.
The square of the area of the triangle is equal to , which is a quadratic function. The graph of this function will be a parabola opening downward, so we can write the maximum value of this function as:
We can then write as and calculate the minimum area of the shaded region. Substituting this value into the area equation, we find its minimum area: .
E) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to .
The substitution will transform this into the quadratic function
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value into the equation gives us and consequently . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to
Substituting the value into the equation (2), we find that the minimum value of the shaded region is equal to .
. Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle.A) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to
.The square of the area of the triangle is equal to
, and the substitution will transform this expression into the quadratic function Since we want to find the maximum value of t, we will substitute the value
into the equation. Solving for t gives us the following minimum area of the shaded region: .B) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to
.The square of the area of the triangle is equal to
The substitution will transform this expression into the quadratic function Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value
into the equation gives us and consequently (The negative root can be rejected since the side of a triangle can't be negative). With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to Substituting the value
in the equation (2) gives us that the minimum value of the shaded region is equal to .C) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to
.The substitution
will transform this into the quadratic function Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value
into the equation gives us and consequently . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to Substituting the value
into the equation (2), we find that the minimum value of the shaded region is equal to .D) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle.
The square of the area of the triangle is equal to
, which is a quadratic function. The graph of this function will be a parabola opening downward, so we can write the maximum value of this function as: We can then write
as and calculate the minimum area of the shaded region. Substituting this value into the area equation, we find its minimum area: .E) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to
.The substitution
will transform this into the quadratic function Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is
Substituting the value
into the equation gives us and consequently . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to Substituting the value
into the equation (2), we find that the minimum value of the shaded region is equal to . Unlock Deck
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7
Plot the following points: (6, 2), (7, 5), (8, 8), (9, 9). In your scatter diagram, sketch a line that best seems to fit the data. Estimate the slope and the y-intercept of the line.
A) Estimated slope: 2; Estimated y-intercept: -12.5
B) Estimated slope: 2.5; Estimated y-intercept: -13
C) Estimated slope: 2.9; Estimated y-intercept: -15.3
D) Estimated slope: 1.2; Estimated y-intercept: -1.5
E) Estimated slope: 2.5; Estimated y-intercept: -16
A) Estimated slope: 2; Estimated y-intercept: -12.5
B) Estimated slope: 2.5; Estimated y-intercept: -13
C) Estimated slope: 2.9; Estimated y-intercept: -15.3
D) Estimated slope: 1.2; Estimated y-intercept: -1.5
E) Estimated slope: 2.5; Estimated y-intercept: -16
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8
Sketch the graph of the function and specify all x- and y-intercepts. y =(x - 2) 2 + 4
A) There is no x-intercept. y-intercept: 8
B) There is no x-intercept. y-intercept: 8
C) There is no x-intercept. y-intercept: 7
D) x-intercept: 0 y-intercept: 4
E) There is no x-intercept. y-intercept: 7
A)
There is no x-intercept. y-intercept: 8B)
There is no x-intercept. y-intercept: 8C)
There is no x-intercept. y-intercept: 7D)
x-intercept: 0 y-intercept: 4E)
There is no x-intercept. y-intercept: 7 Unlock Deck
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9
What is the largest possible area for a rectangle with a perimeter of 40 cm?
A) 400 cm 2
B) 100 cm 2
C) 90 cm 2
D) 130 cm 2
E) 150 cm 2
A) 400 cm 2
B) 100 cm 2
C) 90 cm 2
D) 130 cm 2
E) 150 cm 2
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10
Find the linear function satisfying the given conditions. and
A)
B)
C)
D)
E)
and A)
B)
C)
D)
E)
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11
For the following figure, express the length AB as a function of x. (Hint: Note the similar triangles.)
A)
B)
C)
D)
E)
A)
B)
C)
D)
E)
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12
Find all fixed points of the function.
A) x = 1
B) x = - 1
C) x = 4
D) x = 3
E) no fixed points
A) x = 1
B) x = - 1
C) x = 4
D) x = 3
E) no fixed points
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13
A factory owner buys a new machine for $25,000. After eight years, the machine has a salvage value of $1,000. Find a formula for the value of the machine after t years, where
A) V (t) = - 3,000x + 25,000
B) V (t) = - 3,000x + 1,000
C) V (t) = 3,000x + 25,000
D) V (t) = 3,000x - 25,000
E) V (t) = - 3,000x - 25,000
A) V (t) = - 3,000x + 25,000
B) V (t) = - 3,000x + 1,000
C) V (t) = 3,000x + 25,000
D) V (t) = 3,000x - 25,000
E) V (t) = - 3,000x - 25,000
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14
Sketch the graph of the function and specify all x- and y-intercepts. y = (x - 2)(x - 1)(x + 1)
A) x-intercepts: - 2, 1, 2
Y-intercept: -4
B) x-intercepts: - 1, 1, 2
Y-intercept: 2
C) x-intercepts: - 1, 1, 2
Y-intercept: -2
D) x-intercepts: - 2, 1, 2
Y-intercept: 4
E) x-intercepts: - 1, 1
Y-intercepts: - 2.25
A)
x-intercepts: - 2, 1, 2Y-intercept: -4
B)
x-intercepts: - 1, 1, 2Y-intercept: 2
C)
x-intercepts: - 1, 1, 2Y-intercept: -2
D)
x-intercepts: - 2, 1, 2Y-intercept: 4
E)
x-intercepts: - 1, 1Y-intercepts: - 2.25
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15
Two points A and B move along the x-axis. After t sec, their positions are given by the equations At what time t do A and B have the same x-coordinate?
A) 7 sec
B) 6 sec
C) 5 sec
At what time t do A and B have the same x-coordinate?A) 7 sec
B) 6 sec
C) 5 sec
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16
Find all fixed points of the function.
A) x = - 1, x = - 9
B) x = - 1, x = 9
C) x = - 9, x = 1
D) x = 9
E) no fixed points
A) x = - 1, x = - 9
B) x = - 1, x = 9
C) x = - 9, x = 1
D) x = 9
E) no fixed points
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17
Determine the input that produces the largest or smallest output (whichever is appropriate). State whether the output is largest or smallest. s = - 24t 2 + 147t + 15
A) ; largest
B) ; largest
C) ; smallest
D) ; largest
E) ; smallest
A)
; largestB)
; largestC)
; smallestD)
; largestE)
; smallest Unlock Deck
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18
A piece of wire inches long is bent into a circle. Express the area of the circle as a function of y.
A)
B)
C)
D)
E)
inches long is bent into a circle. Express the area of the circle as a function of y.A)
B)
C)
D)
E)
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19
Suppose that the height of an object shot straight up is given by h =544t - 16t 2 . (Here h is in feet and t is in seconds.) Find the maximum height.
A) 4,724 ft
B) 4,624 ft
C) 1,156 ft
D) 4,290 ft
E) 5,189 ft
A) 4,724 ft
B) 4,624 ft
C) 1,156 ft
D) 4,290 ft
E) 5,189 ft
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20
Among all rectangles having a perimeter of 24m, find the dimensions of the one with the largest area.
A)
B)
C)
D)
E)
A)
B)
C)
D)
E)
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21
Sketch the graph of the rational function. Specify the intercepts and the asymptotes.
A) no x-intercepts; y-intercept: 1; no vertical asymptotes; horizontal asymptote: y = 0;
B) no x-intercepts; y-intercept: 1; vertical asymptote: x = 1; horizontal asymptote: y = 0;
C) no x-intercepts; y-intercept: 1; vertical asymptote: x = - 1; horizontal asymptote: y = 0;
D) no x-intercepts; y-intercept: -1; vertical asymptote: x = 1; horizontal asymptote: y = 0;
E) no x-intercepts; y-intercept: -1; no vertical asymptotes; horizontal asymptote: y = 0;
A) no x-intercepts; y-intercept: 1; no vertical asymptotes; horizontal asymptote: y = 0;
B) no x-intercepts; y-intercept: 1; vertical asymptote: x = 1; horizontal asymptote: y = 0;
C) no x-intercepts; y-intercept: 1; vertical asymptote: x = - 1; horizontal asymptote: y = 0;
D) no x-intercepts; y-intercept: -1; vertical asymptote: x = 1; horizontal asymptote: y = 0;
E) no x-intercepts; y-intercept: -1; no vertical asymptotes; horizontal asymptote: y = 0;
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22
Sketch the graph of the function and specify all x- and y-intercepts.y = x 3 - 9x
A) x-intercept: 0
Y-intercept: 0
B) x-intercept: 0
Y-intercept: 0
C) x-intercepts: -3, 0, 3
Y-intercept: 0
D) x-intercepts: -3, 0, 3
Y-intercept: 0
E) x-intercept: 0
Y-intercept: 0
A)
x-intercept: 0Y-intercept: 0
B)
x-intercept: 0Y-intercept: 0
C)
x-intercepts: -3, 0, 3Y-intercept: 0
D)
x-intercepts: -3, 0, 3Y-intercept: 0
E)
x-intercept: 0Y-intercept: 0
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23
Sketch the graph of the rational function. Specify the intercepts and the asymptotes.
A) x-intercept: 3; y-intercept: - 1; vertical asymptote: x = - 3; horizontal asymptote: y = 1;
B) x-intercept: 0; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = - 1;
C) x-intercept: 3; y-intercept: 1; vertical asymptote: x = - 3; horizontal asymptote: y = - 1;
D) x-intercept: 3; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = - 1;
E) x-intercept: - 3; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = 1;
A) x-intercept: 3; y-intercept: - 1; vertical asymptote: x = - 3; horizontal asymptote: y = 1;
B) x-intercept: 0; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = - 1;
C) x-intercept: 3; y-intercept: 1; vertical asymptote: x = - 3; horizontal asymptote: y = - 1;
D) x-intercept: 3; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = - 1;
E) x-intercept: - 3; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = 1;
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24
Sketch the graph of the function and specify all x- and y-intercepts. y = x 3 (x + 1)
A) x-intercepts: 0, 1
Y-intercept: 0
B) x-intercepts: 0, - 1
Y-intercept: 0
C) x-intercepts: 0, 1
Y-intercept: 0
D) x-intercepts: 0, 1
Y-intercept: 0
E) x-intercepts: 0, - 1
Y-intercept: 0
A)
x-intercepts: 0, 1Y-intercept: 0
B)
x-intercepts: 0, - 1Y-intercept: 0
C)
x-intercepts: 0, 1Y-intercept: 0
D)
x-intercepts: 0, 1Y-intercept: 0
E)
x-intercepts: 0, - 1Y-intercept: 0
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25
Sketch the graph of the rational function. Specify the intercepts and the asymptotes.
A) no x-intercepts; y-intercept: ; vertical asymptote: x = - 4; horizontal asymptote: y = 0;
B) no x-intercepts; y-intercept: ; vertical asymptote: x = 4; horizontal asymptote: y = 0;
C) no x-intercepts; y-intercept: ; vertical asymptote: x = 4; horizontal asymptote: y = 0;
D) no x-intercepts; y-intercept: ; vertical asymptote: x = 4; horizontal asymptote: y = 0;
E) no x-intercepts; y-intercept: ; vertical asymptote: x = 4; horizontal asymptote: y = 0;
A) no x-intercepts; y-intercept:
; vertical asymptote: x = - 4; horizontal asymptote: y = 0; B) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; C) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; D) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; E) no x-intercepts; y-intercept:
; vertical asymptote: x = 4; horizontal asymptote: y = 0; Unlock Deck
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