Table of Contents
Chapter 1: Systems of Linear Equations 2
Getting Started 1.1 Representing Linear Relations 1.2 Solving Linear Equations 1.3 Graphically Solving Linear Systems
Curious Math
4 8 15 21 29 30 33 41 49 57 60 64 65
Chapter 3: Graphs of Quadratic Relations
Getting Started 3.1 Exploring Quadratic Relations
Curious Math
128
130 134 149 150 159 161 169 179 183 187 188 189
3.2 Properties of Graphs of Quadratic Relations 138 3.3 Factored Form of a Quadratic Relation Mid-Chapter Review 3.4 Expanding Quadratic Expressions 3.5 Quadratic Models Using Factored Form 3.6 Exploring Quadratic and Exponential Graphs Chapter Review Chapter Self-Test Chapter Task Chapters 1–3 Cumulative Review
Mid-Chapter Review 1.4 Solving Linear Systems: Substitution 1.5 Equivalent Linear Systems 1.6 Solving Linear Systems: Elimination 1.7 Exploring Linear Systems Chapter Review Chapter Self-Test Chapter Task
Chapter 2: Analytic Geometry: Line Segments and Circles
Getting Started 2.1 Midpoint of a Line Segment 2.2 Length of a Line Segment 2.3 Equation of a Circle Mid-Chapter Review 2.4 Classifying Figures on a Coordinate Grid 2.5 Verifying Properties of Geometric Figures 2.6 Exploring Properties of Geometric Figures
Curious Math
66
68 72 81 88 94 96 104 111 114 115 122 126 127
Chapter 4: Factoring Algebraic Expressions
Getting Started 4.1 Common Factors in Polynomials 4.2 Exploring the Factorization of Trinomials 4.3 Factoring Quadratics: Mid-Chapter Review 4.4 Factoring Quadratics: ax 2 + bx + c 4.5 Factoring Quadratics: Special Cases
Curious Math
192
194 198 205 207 214 217 225 232 233 238 242 243
x2
+ bx + c
2.7 Using Coordinates to Solve Problems Chapter Review Chapter Self-Test Chapter Task
4.6 Reasoning about Factoring Polynomials Chapter Review Chapter Self-Test Chapter Task
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Chapter 5: Applying Quadratic Models
Getting Started 5.1 Stretching/Reflecting Quadratic Relations 5.3 Graphing Quadratics in Vertex Form Mid-Chapter Review 5.4 Quadratic Models Using Vertex Form
Curious Math
7.2 Solving Similar Triangle Problems
382 389 391 394 400 407 408 415 418 419
244
246 250 263 273 275 296 297 303 306 307
Mid-Chapter Review 7.3 Exploring Similar Right Triangles 7.4 The Primary Trigonometric Ratios 7.5 Solving Right Triangles
Curious Math
5.2 Exploring Translations of Quadratic Relations 259
7.6 Solving Right Triangle Problems Chapter Review Chapter Self-Test Chapter Task
5.5 Solving Problems Using Quadratic Relations 285 5.6 Connecting Standard and Vertex Forms Chapter Review Chapter Self-Test Chapter Task
Chapter 8: Acute Triangle Trigonometry
Getting Started 8.1 Exploring the Sine Law 8.2 Applying the Sine Law Mid-Chapter Review 8.3 Exploring the Cosine Law
Curious Math
420
422 426 428 435 437 439 440 446 452 454 455 456
Chapter 6: Quadratic Equations
Getting Started 6.1 Solving Quadratic Equations 6.2 Exploring the Creation of Perfect Squares
Curious Math
308
310 314 322 324 325 333 336 345 352 360 363 364 365
8.4 Applying the Cosine Law 8.5 Solving Problems Using Acute Triangles Chapter Review Chapter Self-Test Chapter Task Chapters 7–8 Cumulative Review Appendix A: REVIEW OF ESSENTIAL SKILLS AND KNOWLEDGE Appendix B: REVIEW OF TECHNICAL SKILLS Glossary
6.3 Completing the Square Mid-Chapter Review 6.4 The Quadratic Formula 6.5 Interpreting Quadratic Equation Roots 6.6 Solving Problems Using Quadratic Models Chapter Review Chapter Self-Test Chapter Task Chapters 4–6 Cumulative Review
459 486 518 525 584 588
Chapter 7: Similar Triangles and Trigonometry
Getting Started 7.1 Congruence and Similarity in Triangles
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370 374
Answers Index Credits
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Chapter
5
Applying Quadratic Models
GOALS
You will be able to
• • • •
Investigate the y a(x a quadratic relation
h)2
k form of
Apply transformations to sketch graphs of quadratic relations Apply quadratic models to solve problems Investigate connections among the different forms of a quadratic relation
An arch is a structure that spans a distance and supports weight. The ancient Romans were the first people to use semicircular arches in a wide range of structures. The arches in this bridge are the strongest type.
? What characteristics of these arches
suggest that they are parabolas?
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5
a) b) c) d) e) f) transformation translation reflection parabola vertex factored form of a quadratic relation
Getting Started
WORDS YOU NEED to Know
1. Match each term at the left with the correct description or example.
i) a point that relates to the maximum or minimum value of a quadratic relation ii) the result of moving or changing the size of a shape according to a rule iii) the result of sliding each point on a shape the same distance in the same direction iv) y = a(x - r)(x - s) v) the result of flipping a shape to produce a mirror image of the shape vi) the graph of a quadratic relation
Study
Aid
SKILLS AND CONCEPTS You Need
Working with Transformations
Translations, reflections, rotations, and dilatations are types of transformations. They can be applied to a point, a line, or a shape.
EXAMPLE
• For more help and practice,
see Appendix A-13.
6 4 2 A -4 -2
0
y C B 2 4 6 x
-2 -4 6 4 2 A y
Apply the following transformations to ^ ABC shown at the left. a) Translate ^ ABC 3 units right and 2 units up. b) Reflect ^ ABC in the x-axis.
Solution
C A′ B B″ 4 C″
C′ B′ x 6
Apply the same translation and the same reflection to points A, B, and C. Plot the image points, and draw each image triangle. a)
Original Point A(0, 1) B(2, 1) C(2, 4) Image Point A¿ (3, 3) B¿ (5, 3) C¿ (5, 6)
b)
Original Point A(0, 1) B(2, 1) C(2, 4)
Image Point A– (0, B– (2, C– (2, y 1) 1) 4)
-4
-2 A″ -2 -4
0
2. a) In the diagram at the right, is figure B,
figure C, or figure D the result of a translation of figure A? Explain. b) Which figure is the result of a reflection of figure A in the x-axis? Explain. c) Which figure is the result of a reflection of figure A in the y-axis? Explain.
246
Getting Started Draft Evaluation Copy
B 2
A x
-4
-2
0
-2 -4 D
4
C
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Getting Started
Understanding Quadratic Relations
A quadratic relation can be expressed algebraically as an equation in standard form or factored form. You can determine information about its parabola from the equation of the relation.
EXAMPLE
Study
Aid
• For more help and practice, see Lesson 3.3, Examples 2 to 4.
Determine the properties of the relation defined by the equation y = (x - 2)(x - 4). Then sketch the graph of the relation.
Solution
The equation y = (x - 2)(x - 4) is the equation of a quadratic relation in factored form. The values x = 2 and x = 4 make the factors (x - 2) and (x - 4) equal to zero. These are called the zeros of the relation and are the x-intercepts of the graph. The axis of symmetry is the perpendicular bisector of the line segment 2 + 4 that joins the zeros. Its equation is x = or x = 3. 2 The vertex of the parabola lies on the axis of symmetry. To determine the y-coordinate of the vertex, substitute x = 3 into the equation and evaluate. y = (3 - 2)(3 - 4) y = (1)( -1) y = -1 The vertex is (3, -1).
y 6 4 2
0
(x x 3
2)(x
4)
y
x-intercepts x
-2
2
4 6 vertex
8
3. Determine the value of y in each quadratic relation for each value of x.
a) y = x 2 + 2x + 5, when x = -4 b) y = x 2 - 3x - 28, when x = 7
4. Determine the zeros, the equation of the axis of symmetry, and the
vertex of each quadratic relation. a) y = (x + 5)(x - 3) b) y = 2(x - 4)(x + 1)
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c)
y = -4x(x + 3)
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PRACTICE
Study
Aid
5. State the coordinates of the image point after applying the indicated
• For help, see the Review of
Essential Skills and Knowledge Appendix.
Question 5, 6 y
Appendix A-13
transformation(s). a) A(3, 4) is translated 2 units left and 5 units up. b) B( 1, 5) is translated 4 units right and 3 units down. c) C(2, 7) is translated 2 units left and 7 units up. d) D(3, 5) is reflected in the x-axis.
6. Describe the transformation that was used to translate each triangle
8
6 Q P R Q′ 2
0
onto the image in the diagram at the left. a) ^PQR to ^P¿Q¿R¿ b) ^PQR to ^P–Q–R– c) ^P¿Q¿R¿ to ^P–Q–R–
R′ x 6 8 R″ 7. State the zeros, equation of the axis of symmetry, and the vertex
P′ Q″ P″
of each quadratic relation. a)
4 2 x -4 -2
0
-2
-2 -4
y
b)
4 2
y
x -6 -4 -2
0
-2 -4 -6 -8
2
4
-2 -4 -6 -8
2
-10
-10
8. Express each quadratic relation in standard form.
a) y = (x + 4)(x + 5) b) y = (2x - 3)(x + 2)
c) y = -3(x - 4)(x + 7) d) y = (x + 5)2
9. Sketch the graph of each quadratic relation in question 8. 10. Express y = 2x 2 - 4x - 48 in factored form. Then determine
its minimum value.
11. Copy and complete the chart to show what you know about quadratic
relations. Share your chart with a classmate.
Definition: Quadratic Relation Special Properties:
Examples:
Non-examples:
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Getting Started
APPLYING What You Know
Tiling Transformers
Jesse and Tyler decided to have a competition to see who could transform the figures from the Start grid to the End grid in the fewest number of moves.
Start 8 6 4 -8 -6 -4 x -8 -6 ? A. B. C. D. E. -4 -2
0 0
YOU WILL NEED • • • •
grid paper ruler coloured pencils or markers scissors
y
End y 4 x 4 6 8
-2
2
-4
What is the fewest number of moves needed to transform these figures?
On a piece of grid paper, draw the five figures on the Start grid. Cut out each figure. Draw x- and y-axes on another piece of grid paper. Label each axis from 10 to 10. Put the yellow figure on the grid paper in the position shown on the Start grid. Apply one or more transformations to move the yellow figure to the position shown on the End grid. Make a table like the one below. Record the transformation(s) that you used to move the yellow figure for part D.
Player 1 Original Coordinates New Coordinates A(–6, 4) -8 -6 -4 A′(–1, –2) -2 -4 B(–3, 7) 8 6 4 2 x
0
y
B′(2, 1)
2
4
Move
Figure
Transformation translation 5 units right, 6 units down
1 2 F. G. H.
yellow
A(–6, 4), B(–3, 7)
A (–1, –2), B (2, 1)
Repeat parts C to E for another figure. (The second figure can move across the first figure if necessary.) Continue transforming figures and recording results until you have created the design shown on the End grid. How many moves did you need? Compare your results with those of other classmates. What is the fewest number of moves needed?
Draft Evaluation Copy Chapter 5
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5.1
YOU WILL NEED • graphing calculator • dynamic geometry software
(optional)
Stretching/Reflecting Quadratic Relations
GOAL
Examine the effect of the parameter a in the equation y ax 2 on the graph of the equation.
• grid paper • ruler
INVESTIGATE the Math
Suzanne’s mother checks the family’s investments regularly. When Suzanne saw the stock chart that her mother was checking, she noticed trends in sections of the graph. These trends looked like the shapes of the parabolas she had been studying. Each “parabola” was a different shape.
As of 10-Jun-09 80.00 Canadian Dollars ? A. 75.00 70.00 65.00 60.00
What is the relationship between the value of a in the equation y ax 2 and the shape of the graph of the relation?
Enter y = x 2 as Y1 in the equation editor of a graphing calculator. The window settings shown are “friendly” because they allow you to trace using intervals of 0.1. Graph the parabola using these settings. Enter y = 2x 2 in Y2 and y = 5x 2 in Y3, and graph these quadratic relations. What appears to be happening to the shape of the graph as the value of a increases?
Tech
Support
B.
For help graphing relations and adjusting the window settings, see Appendix B-2 and B-4.
C.
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5.1
D.
E.
Where would you expect the graph of y = 3x 2 to appear, relative to the other three graphs? Check by entering y = 3x 2 into Y4 and graph with a bold line. Was your conjecture correct? 1 1 Where would you expect the graphs of y = x 2 and y = x 2 to 2 4 2? Clear the equations from Y2, appear, relative to the graph of y = x 1 1 Y3, and Y4. Enter y = x 2 into Y2 and y = x 2 into Y3, and graph 2 4 these quadratic relations. Describe the effect of the parameter a on the parabola when 0 a 1. 3 2 x to appear, relative to 4 3 the other three graphs? Check by entering y = x 2 into Y4 and graph 4 with a bold line. Where you would expect the graph of y = Clear the equations from Y2, Y3, and Y4. Enter y = -4x 2 into Y2 1 and y = - x 2 into Y3, and graph these quadratic relations. Describe 4 the effect of a on the parabola when a 0. Ask a classmate to give you an equation in the form y = ax 2, where a 0. Describe to your classmate what its graph would look like relative to the other three graphs. Verify your description by graphing the equation in Y4. How does changing the value of a in the equation y = ax 2 affect the shape of the graph?
Tech
Support
Move the cursor to the left of Y4. Press ENTER to change the = line style to bold.
parameter a coefficient that can be changed in a relation; for example, a, b, and c are parameters in y ax2 bx
F.
c
G.
H.
I.
Reflecting
J.
Which parabola in the stock chart has the greatest value of a? Which has the least value of a? Which parabolas have negative values of a? Explain how you know. What happens to the x-coordinates of all the points on the graph of y = x 2 when the parameter a is changed in y = ax 2? What happens to the y-coordinates? What happens to the shape of the parabola near its vertex? State the ranges of values of a that will cause the graph of y = x 2 to be i) vertically stretched ii) vertically compressed iii) reflected across the x-axis
vertical stretch a transformation that increases all the y-coordinates of a relation by the same factor vertical compression a transformation that decreases all the y-coordinates of a relation by the same factor
K.
L.
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APPLY the Math
EXAMPLE
1
Selecting a transformation strategy to graph a parabola
a) Sketch the graph of the equation y = 3x 2 by transforming the graph of y = x 2. b) Describe how the graphs of y = 3x 2 and y = -3x 2 are related. Zack’s Solution a)
x y 2 4 1 1 12 10 8 6 4 2 x -8 -6 -4 -2
0
0 0 y
1 1
2 4
I created a table of values to determine five points on the graph of y x2.
y
x2 I plotted the points on a grid and joined them with a smooth curve. I can use these five points any time I want to sketch the graph of y x2 because they include the vertex and two points on each side of the parabola. I decided to call this my five-point sketch. To transform my graph into a graph of y 3x2, I multiplied the y-coordinates of each point on y x2 by 3. For example, (2, 4) (2, 12) 4 3
-2 -4
2
4
6
8
x y
2 12
1 3
0 0 y
1 3
2 12
y y
x2 3x2
12 10 8 6 4 2
(2, 12)
(2, 4) x
-8 -6 -4
-2
0
-2 -4
2
4
6
8
I plotted and joined my new points to get the graph of y 3x2. a 3 represents a vertical stretch by a factor of 3. This means that the y-coordinates of the points on the graph of y 3x2 will become greater faster, so the parabola will be narrower near its vertex compared to the graph of y x2.
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5.1 Stretching/Reflecting Quadratic Relations
Draft Evaluation Copy
5.1
b)
x y
2 12 12 10 8 6 4 2
1 3 y
0 0
1 3
2 12 To get the graph of y 3x2, I multiplied the y-coordinates of all the points on the graph of y 3x2 by 1. For example,
y
3x2
(2,12) 12
(2, 12) ( 1)
x -4 -2
0
-2 -4 -6 -8
2 y
4 3x2
a 3 represents a vertical stretch by a factor of 3 and a reflection in the x-axis. This means that all the points on the graph of y 3x2 are reflected in the x-axis.
-10 -12 (2, -12)
The graph of y = -3x 2 is the reflection of the graph of y = 3x 2 in the x-axis.
EXAMPLE
2
Determining the value of a from a graph
Determine an equation of a quadratic relation that models the arch of San Francisco’s Bay Bridge in the photograph below.
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Mary’s Solution: Representing the picture on a hand-drawn grid
2 I used a photocopy of the photograph. I laid a transparent grid with axes on top of the photocopy. I placed the origin at the vertex of the arch. I did this since all parabolas defined by y ax2 have their vertex at (0, 0).
-3 -2
5
I located a point on the graph and estimated the coordinates of the point to be (5, 1). y 1 1 1 25 1 25 = ax 2 = a(5)2 = 25a 25a = 25 = a
The equation of the graph is in the form y ax2. To determine the value of a, I had to determine the coordinates of a point on the parabola. I chose the point (5, 1). I substituted x 5 and y 1 into the equation and solved for a.
An equation that models the arch of the bridge 1 2 is y = x . 25
The graph that models the arch is a vertical compression of the graph of y x2 by a factor 1 of . 25
Sandeep’s Solution: Selecting dynamic geometry software
4 2 -3 -2 -4 5
I imported the photograph into dynamic geometry software. I superimposed a grid over the photograph. Then I adjusted the grid so that the origin was at the vertex of the bridge’s parabolic arch. I need to create a graph using the relation y = ax2 by choosing a value for the parameter a. Tech
Support
For help creating and graphing relations using parameters in dynamic geometry software, as well as animating the parameter, see Appendix B-17.
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5.1 Stretching/Reflecting Quadratic Relations
Draft Evaluation Copy
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5.1
f (x) = a •x2 a = 1.00
4 2 -3 -2 -4 5
When I used a 1, the graph of y = x 2 appeared. The parabola was too narrow. It had to be vertically compressed to fit the arch. To do this, I needed a lower value of a, between 0 and 1. I needed a positive value because the arch opens upward.
f (x) = a •x2 a = 0.04
4 2 -3 -2 -4 5
I tried a enough.
0.5, but the parabola was not wide
I tried a 0.1. This value gave me a better fit. I still wasn’t satisfied, so I tried different values of a 0.04 gave me between 0 and 0.1. I found that a a good fit. Vertically compressing the graph of y = x 2 by a factor of 0.04 creates a graph that fits the photograph.
An equation that models the bridge is y = 0.04x 2.
In Summary
Key Idea
• When compared with the graph of y x2, the graph of y ax2 is a parabola that has been stretched or compressed vertically by a factor of a.
Need to Know
• Vertical stretches are determined by the value of a. When a 1, the 1, the graph is stretched graph is stretched vertically. When a vertically and reflected across the x-axis. • Vertical compressions are also determined by the value of a. When 0 a 1, the graph is compressed vertically. When 1 a 0, the graph is compressed vertically and reflected across the x-axis. • If a 0, the parabola opens upward. • If a 0, the parabola opens downward. y
ax2, a 1 y
y 0
ax2, a 1
y
x2
x
y 1
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ax2, a 0
y a
ax2, 1
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CHECK Your Understanding
1. Match each graph with the correct equation. The graph of y = x2
is in green in each diagram. a) y = 4x 2 b) y = -3x 2 i)
10 8 6 4 y -4 x2 2 -2
0
2 2 x 3 d) y = -0.4x 2 c) y = iii)
y x2 6 4 2 x -4 -2
0
y
ii)
x2 -2
6 4 y -4 2
y
y
iv)
10 8 6
y
x
0
-2 -4 -6
2
4
-2 -4 -6
2
4
y
x2
4 2 x
0
x 2 4
-2
-4
-2
-2
2
4
2. The graph of y = x 2 is transformed to y = ax 2 (a Z 1). For each
point on y = x 2, determine the coordinates of the transformed point for the indicated value of a. a) (1, 1) when a = 5 c) (5, 25) when a = -0.6 1 b) ( 2, 4), when a = -3 d) ( 4, 16) when a = 2 3. Write the equations of two different quadratic relations that match each description. a) The graph is narrower than the graph of y = x 2 near its vertex. b) The graph is wider than the graph of y = -x 2 near its vertex. c) The graph opens downward and is narrower than the graph of y = 3x 2 near its vertex.
PRACTISING
4. Sketch the graph of each equation by applying a transformation
K
to the graph of y = x 2. Use a separate grid for each equation, and start by sketching the graph of y = x 2. 1 a) y = 3x 2 d) y = x 2 4 3 b) y = -0.5x 2 e) y = - x 2 2 c) y = -2x 2 f ) y = 5x 2
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5.1
5. Describe the transformation(s) that were applied to the graph of y = x 2
to obtain each black graph. Write the equation of the black graph. a)
y x2 14 12 10 8 6 4 2 x -8 -6 -4 -2
0
y
c)
y x2 4 2
y
x -8 -6 -4 -2
0
-2 -4 -6 -8
2
4
6
8
2 y
4
6
8
-10
b)
y x2
8 6 4 2
d)
10 8 6 4
y
x -8 -6 -4 -2
0
y
x2 -2
2 x
0
-2 -4 -6
2
4
6
8 -8 -6 -4 -2 -4 2 4 6 8
6. Andy modelled the arch of the bridge in the photograph at the right
C
y y 2
x2
by tracing a parabola onto a grid. Now he wants to determine the equation of the parabola. Explain the steps he should use to do this, and state the equation.
-4 -2
x
0
7. Determine the equation of a quadratic model for each natural arch. y x -4 -2
0
2
4
y x -6 -4 -2
0
-2 -4
2
4
6
2
4
-2 -4 -6
-2 -4 -6
a) Lover’s Arch on the Isle of Capri in Italy
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b) Jett Arch in Arizona
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8. Identify the transformation(s) that must be applied to the graph
of y = x 2 to create a graph of each equation. Then state the coordinates of the image of the point (2, 4). a) y = 4x 2 c) y = 0.25x 2 e) y = -x 2 2 1 b) y = - x 2 d) y = -5x 2 f) y = x 2 3 5
9. By tracing the bridge at the left onto a grid, determine an equation that
A
models the lower outline of the Sydney Harbour Bridge in Australia. form y = ax 2 will never result in a parabola that is congruent to the parabola y = x 2. Do you agree or disagree? Justify your decision.
10. Seth claims that changing the value of a in quadratic relations of the
T
11. Copy and complete the following table. Direction of Opening (upward/ downward) Description of Transformation (stretch/ compress) Shape of Graph Compared with Graph of y x2 (wider/narrower)
Equation y = 5x 2 y = 0.25x 2 y = 1 2 x 3
y = -8x 2 12. Explain why it makes sense that each statement about the graph
of y = ax 2 is true. a) If a 0, then the parabola opens downward. b) If a is a rational number between 1 and 1, then the parabola is wider than the graph of y = x 2. c) The vertex is always (0, 0).
Extending
13. The graph of y = ax 2 (a Z 1, a
0) is either a vertical stretch or a vertical compression of the graph of y = x 2. Use graphing technology to determine whether changing the value of a has a similar effect on the 1 graphs of equations such as y = ax, y = ax 3, y = ax 4, and y = ax 2. x 2 + y 2 = r 2. a) Explore the effect of changing positive values of a when graphing ax 2 + ay 2 = r 2. b) Explore the effects of changing positive values of a and b when graphing ax 2 + by 2 = r 2.
14. The equation of a circle with radius r and centre (0, 0) is
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5.1 Stretching/Reflecting Quadratic Relations
Draft Evaluation Copy
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5.2
GOAL
Exploring Translations of Quadratic Relations
YOU WILL NEED
Investigate the roles of h and k in the graphs of y y (x h) 2, and y (x h) 2 k.
x2
k,
• grid paper • ruler • graphing calculator
EXPLORE the Math
Hammad has been asked to paint a mural of overlapping parabolas on a wall in his school. A sketch of his final design is shown at the right. He is using his graphing calculator to try to duplicate his design. His design uses parabolas that have the same shape as y = x 2, but he doesn’t know what equations he should enter into his graphing calculator to place the parabolas in different locations on the screen.
? 8 6 4 2 x -8 -6 -4 -2
0
-2 -4 -6 -8
2
4
6
8
What is the connection between the location of the vertex of a parabola and the equation of its quadratic relation?
A.
Enter the equation y = x 2 as Y1 in the equation editor of a graphing calculator. Graph the equation using the window settings shown. Enter an equation of the form y = x 2 + k in Y2 by adding or subtracting a number after the x 2 term. For example, y = x 2 + 1 or y = x 2 - 3. Graph your equation, and compare the graph with the graph of y = x 2. Try several other equations, replacing the one you have in Y2 each time. Be sure to change the number you add or subtract after the x 2 term. Copy this table. Use the table to record your findings for part B.
Equation y x2 Distance and Direction from y not applicable x2 Vertex (0, 0)
Tech
Support
For help graphing relations, changing window settings, and tracing along a graph, see Appendix B-2 and B-4.
B.
Tech
Support
Use the TRACE key and the up arrow to help you distinguish one graph from another.
^
C.
Value of k 0
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D.
Investigate what happens to the graph of y = x 2 when a number is added to or subtracted from the value of x before it is squared, creating an equation of the form y = (x - h)2. For example, y = (x + 1)2 or y = (x - 2)2. Graph your new equations in Y2 each time using a graphing calculator. Then copy this table and record your findings.
Equation y x2 Distance and Direction from y not applicable x2 Vertex (0, 0)
Value of h 0
E.
Identify the type of transformations that have been applied to the graph of y = x 2 to obtain the graphs in your table for part C and your table for part D. Make a conjecture about how you could predict the equation of a parabola if you knew the translations that were applied to the graph of y = x 2. Copy and complete this table to investigate and test your conjecture for part F.
Relationship to y Equation y = x2 Left/Right x2 Vertex (0, 0)
F.
G.
Value Value of h of k 0 0
Up/Down
not applicable not applicable left 3 down 5
4
1
y = (x - 4)2 + 1 ( 2, 6) y = (x + 5)2 - 3
H.
Use what you have discovered to identify the equations that Hammad should type into his calculator to graph the parabolas in the mural design. If the equation of a quadratic relation is given in the form y = (x - h)2 + k, what can you conclude about its vertex?
I.
Reflecting
J.
Describe how changing the value of k in y = x 2 + k affects i) the graph of y = x 2 ii) the coordinates of each point on the parabola y = x 2 iii) the parabola’s vertex and axis of symmetry Describe how changing the value of h in y = (x - h)2 affects i) the graph of y = x 2 ii) the coordinates of each point on the parabola y = x 2 iii) the parabola’s vertex and axis of symmetry
Draft Evaluation Copy
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K.
260
5.2 Exploring Translations of Quadratic Relations
5.2
L.
For parabolas defined by y = (x - h)2 + k, i) how do their shapes compare to the parabola defined by y = x 2? ii) what is the equation of the axis of symmetry? iii) what are the coordinates of the vertex?
In Summary
Key Ideas
• The graph of y = (x - h)2 + k is congruent to the graph of y = x 2, but translated horizontally and vertically. • Translations can also be described as shifts. Vertical shifts are up or down, and horizontal shifts are left or right.
Need to Know
• The value of h tells how far and in what direction the parabola is translated horizontally. If h 6 0, the parabola is translated h units left. If h 7 0, the parabola is translated h units right. • The vertex of y = (x - h)2 is the point (h, 0). • The equation of the axis of symmetry of y = (x - h)2 is x = h. y (x h)2 y h 0
x h 0 y h x2 k y k 0 x k y x2 0 y y (x h)2 k 0
• The value of k tells how far and in what direction the parabola is translated vertically. If k 6 0, the parabola is translated k units down. If k 7 0, the parabola is translated k units up. • The vertex of y = x 2 + k is the point (0, k). • The equation of the axis of symmetry of y = x 2 + k is x = 0.
k
0
• The vertex of y = (x - h)2 + k is the point (h, k). • The equation of the axis of symmetry of y = (x - h)2 + k is x = h.
(h, k) x x h
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FURTHER Your Understanding
1. The following transformations are applied to a parabola with
the equation y = x 2. Determine the values of h and k, and write the equation in the form y = (x - h)2 + k. a) The parabola moves 3 units right. b) The parabola moves 4 units down. c) The parabola moves 2 units left. d) The parabola moves 5 units up. e) The parabola moves 7 units down and 6 units left. f ) The parabola moves 2 units right and 5 units up.
2. Match each equation with the correct graph.
a) y = (x - 2)2 + 3 b) y = (x + 2)2 - 3 i)
8 6 4 2 x -8 -6 -4 -2
0
c) y = (x + 3)2 - 2 d) y = (x - 3)2 + 2
y
y
iii)
8 6 4 2
v)
6 4 2
y
x x -8 -6 -4 -2
0 0
-2 -4
2
-2 y
2
-4
-2
-2 y
2
4
6
ii)
6 4 2
iv)
8 6 4
vi)
6 4 2
y
x -8 -6 -4 -2
0
2 x -2
0
x -4 -2
0
-2 -4
2 -2 2 4 6 8
-2 -4
2
4
6
3. Sketch the graph of each relation by hand. Start with the graph
of y = x 2, and apply the appropriate transformations. a) y = x 2 - 4 b) y = (x - 3)2 c) y = x 2 + 2 d) y = (x + 5)2 e) y = (x + 1)2 - 2 f ) y = (x - 5)2 + 3
4. Describe the transformations that are applied to the graph of y = x 2
to obtain the graph of each quadratic relation. a) y = x 2 + 5 b) y = (x - 3)2 c) y = -3x 2 d) y = (x + 7)2 1 2 x 2 f ) y = (x + 6)2 + 12 e) y =
5. State the vertex and the axis of symmetry of each parabola in question 4.
262
5.2 Exploring Translations of Quadratic Relations
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5.3
GOAL
Graphing Quadratics in Vertex Form
YOU WILL NEED • grid paper • ruler • graphing calculator
Graph a quadratic relation in the form y = a(x - h)2 + k by hand and with graphing technology.
LEARN ABOUT the Math
Srinithi and Kevin are trying to sketch the graph of the quadratic relation y = 2(x - 3)2 - 8 by hand. They know that they need to apply a series of transformations to the graph of y = x 2.
?
How do you apply transformations to the quadratic relation y x2 to sketch the graph of y = 2(x - 3)2 - 8?
Selecting a transformation strategy to graph a quadratic relation Use transformations to sketch the graph of y = 2(x - 3)2 - 8.
EXAMPLE
1
Srinithi’s Solution: Applying a horizontal translation first
y = x2
x y 2 4 1 1 0 0 1 1 2 4
I began by graphing y = x 2 using five key points. The quadratic relation y = 2(x - 3)2 - 8 is expressed in vertex form. Since h 3, I added 3 to the x-coordinate of each point on y = x 2. This means that the vertex is (3, 0).
vertex form a quadratic relation of the form y = a(x - h)2 + k, where the vertex is (h, k)
y = (x x y 1 4
3)2
2 1 3 0 4 1 5 4
y
x2 8 6 4 2 y
y
(x
3)2 The equation of the new red graph is y = (x - 3)2. To draw it, I translated the green parabola 3 units right. x
-4
-2
0
-2 -4
2
4
6
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y = 2(x - 3)2
x y y 1 8 2(x 8 6 4 2 x -4 -2
0
2 2 3)2 y
3 0 y
4 2 (x
5 8 3)2
Since a = 2, I multiplied all the y-coordinates of the points on the red graph by 2. The vertex stays at (3, 0). The equation of this graph is y = 2(x - 3)2. To draw this new blue graph, I applied a vertical stretch by a factor of 2 to the red graph. The blue graph looks correct because the graph with the greater a value should be narrower than the other graph.
-2 -4
2
4
6
y = 2(x - 3)2 - 8
x y 1 0 2 6 y y 3 8 2(x 4 6 3)2 5 0 8
I knew that k = -8. I subtracted 8 from the y-coordinate of each point on the blue graph. The vertex is now (3, 8). The equation of the graph is y = 2(x - 3)2 - 8.
y
2(x
4 3)2 2 -2
0
x -4 -2 -4 -6 -8 2 4 6
Since k 6 0, I knew that I had to translate the blue graph 8 units down to get the final black graph.
Kevin’s Solution: Applying a vertical stretch first
8 y x2 6 4 2 x -4 -2
0
y y 2x2 Since a 2, I decided to stretch the graph of y = x 2 vertically by a factor of 2. I multiplied the y-coordinate of each point on the graph of y = x 2 by 2. The equation of the resulting red graph is y = 2x 2.
-2 -4
2
4
6
264
5.3 Graphing Quadratics in Vertex Form
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5.3
y 2x2 4 2 x -4 -2
0
y
-2 -4 -6 -8
2
4
6
I applied both translations in one step. Adding 3 to the x-coordinate and subtracting 8 from the y-coordinate from each point on the red graph causes the red graph to move 3 units right and 8 units down. 3)2 8 The equation of the resulting black graph is y = 2(x - 3)2 - 8.
y (3, –8)
2(x
Reflecting
A.
Why was it not necessary for Kevin to use two steps for the translations? In other words, why did he not have to shift the graph to the right in one step, and then down in another step? What are the advantages and disadvantages of each solution? How can thinking about the order of operations applied to the coordinates of points on the graph of y = x 2 help you apply transformations to draw a new graph?
B. C.
APPLY the Math
EXAMPLE
2
Reasoning about sketching the graph of a quadratic relation
Sketch the graph of y = -3(x + 5)2 + 1, and explain your reasoning. Winnie’s Solution: Connecting a sequence of transformations to the equation Applying a vertical stretch of factor 3 and a reflection in the x-axis gives the graph of y = -3x 2.
y y 3x2 6 4 2 x -6 -4 -2
0
In the quadratic relation y = -3(x + 5)2 + 1, the value of a is 3. This represents a vertical stretch by a factor of 3 and a reflection in the x-axis.
y
x2 I noticed that I can combine the stretch and reflection into a single step by multiplying each y-coordinate of points on y = x2 by 3.
-2 -4 -6
2
4
6
y
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3x2
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In the equation, h = -5 and k = 1. Therefore, the vertex is at (-5, 1). I translated the blue graph 5 units left and 1 unit up.
y 6 4 2 x -6 -4 -2
0
I determined that the vertex is ( 5, 1). Then I shifted all the points on the graph of y = -3x 2 so that they were 5 units left and 1 unit up.
y
x2
I drew a smooth curve through the new points to sketch the graph.
-2 -4 -6
2
4
6
y y 3(x 5)2 1
3x2
Beth’s Solution: Connecting the properties of a parabola to the equation Based on the equation y = -3(x + 5)2 + 1, the parabola has these properties: • Since a 6 0, the parabola opens downward. • The vertex of the parabola is ( 5, 1). • The equation of the axis of symmetry is x = -5. y y y y = = = = -3(-3 + 5)2 + 1 -3(2)2 + 1 -12 + 1 -11 11) is a point on the parabola.
2 -8 -6 -4 -2
0
Since the equation was given in vertex form, I listed the properties of the parabola that I could determine from the equation.
To determine another point on the parabola, I let x 3.
Therefore, ( 3,
y x I plotted the vertex and the point I had determined, ( 3, 11). Then I drew the axis of symmetry. I used symmetry to determine the point directly across from ( 3, 11). This point is ( 7, 11). I plotted the points and joined them with a smooth curve.
-2 -4 -6 -8
2
4
6
8
-10 -12 x y 5 3(x 5)2 -14 1
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5.3 Graphing Quadratics in Vertex Form
5.3
EXAMPLE
3
Reasoning about the effects of transformations on a quadratic relation
h 100 80 60 40 20 t -10 h
0
For a high school charity event, the principal pays to drop a watermelon from a height of 100 m. The height, h, in metres, of the watermelon after t seconds is h = -0.5gt 2 + k, where g is the acceleration due to gravity and k is the height from which the watermelon is dropped. On Earth, g = 9.8 m/s2. a) The clock that times the fall of the watermelon runs for 3 s before the principal releases the watermelon. How does this change the graph shown? Determine the equation of the new relation. b) On Mars, g = 3.7 Suppose that an astronaut dropped a watermelon from a height of 100 m on Mars. Determine the equation for the height of the watermelon on Mars. How does the graph for Mars compare with the graph for Earth in part a)? c) The principal drops another watermelon from a height of 50 m on Earth. How does the graph in part a) change? How does the relation change? d) Repeat part c) for an astronaut on Mars. Nadia’s Solution a) The equation of the original relation is h = -0.5(9.8)t 2 + 100 h = -4.9t 2 + 100, where t Ú 0 The parabola is translated 3 units right. The equation of the new relation is h = -4.9(t - 3)2 + 100, where t Ú 3.
h 100 80 60 40 20 t -10 h
0
10 100
m/s2.
4.9t2
The original graph is a parabola that opens downward, with vertex (0, k) (0, 100). I wrote and simplified the original relation. I subtracted 3 from the t-coordinate to determine the new relation. Since the watermelon is not falling before 3 s, the relation only holds for t Ú 3.
If the clock runs for 3 s before the watermelon is dropped, then the watermelon will be at its highest point at 3 s. So, the vertex of the new parabola is (3, 100), which is a shift of the original parabola 3 units right.
10 3)2 100
4.9(t
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b) The equation of the relation on Mars is h = -0.5(3.7)t 2 + 100 h = -1.85t 2 + 100, where t Ú 0 The graph for Mars is wider near the vertex.
h 100 80 60 40 20 t -10 h
0
I used the value of g on Mars, g = 3.7m/s2, instead of g = 9.8 m/s2.
A smaller (negative) a-value means that the parabola is wider.
10 100
1.85t2
The t-intercept is farther from the origin, so the watermelon would take longer to hit the ground on Mars compared to Earth. c) The equation of the new relation is h = -4.9t 2 + 50, where t Ú 0. The new graph has the same shape but is translated 50 units down.
h 60 40 20 t -10 h
0
In the relation, k changes from 100 to 50.
The new vertex is half the distance above the origin, at (0, 50) instead of (0, 100). This is a shift of 50 units down. 10 50
4.9t2
d) The new graph for Mars is wider than the original graph and is translated 50 units down.
h 60 40 20 t -10 h
0
The new graph for Mars is wider than the original graph, like the graph in part b). It is translated down, like the graph in part c).
10 50
1.85t2
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5.3 Graphing Quadratics in Vertex Form
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5.3
In Summary
Key Idea
• Compared with the graph of y = x 2, the graph of y = a(x - h)2 + k is a parabola that has been stretched or compressed vertically by a factor of a, translated horizontally by h, and translated vertically by k. As well, if a 6 0, the parabola is reflected in the x-axis.
Need to Know
• The vertex of y = a(x - h)2 + k has the coordinates (h, k). The equation of the axis of symmetry of y = a(x - h)2 + k is x = h. • When sketching the graph of y = a(x - h)2 + k as a transformation of the graph of y = x 2, follow the order of operations for the arithmetic operations to be performed on the coordinates of each point. Apply vertical stretches/compressions and reflections, which involve multiplication, before translations, which involve addition or subtraction.
y y
a(x
h)2
k
x y x2 y ax2 (h, k) x h
CHECK Your Understanding
1. Describe the transformations you would apply to the graph of y = x 2,
in the order you would apply them, to obtain the graph of each quadratic relation. 1 a) y = x 2 - 3 c) y = - x 2 2 b) y = (x + 5)2 d) y = 4(x + 2)2 - 16
2. For each quadratic relation in question 1, identify
i) the direction in which the parabola opens ii) the coordinates of the vertex iii) the equation of the axis of symmetry
3. Sketch the graph of each quadratic relation. Start with a sketch of y = x 2,
and then apply the appropriate transformations in the correct order. a) y = (x + 5)2 - 4 c) y = 2(x - 3)2 1 b) y = -0.5x 2 + 8 d) y = (x - 4)2 - 2 2
PRACTISING
4. What transformations would you apply to the graph of y = x 2
to create the graph of each relation? List the transformations in the order you would apply them. a) y = -x 2 + 9 c) y = (x + 2)2 - 1 b) y = (x - 3)2 d) y = -x 2 - 6
NEL
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1 e) y = -2(x - 4)2 + 16 g) y = - (x + 4)2 - 7 2 1 2 + 12 f ) y = (x + 6) h) y = 5(x - 4)2 - 12 2 5. Sketch a graph of each quadratic relation in question 4 on a separate grid. Use the properties of the parabola and additional points as necessary.
6. Match each equation with the correct graph.
1 (x - 2)2 - 5 2 1 b) y = (x - 4)2 - 2 2 a) y = c) y = -2(x + 2)2 + 5 i)
8 6 4 2 x -4 -2
0
d) y = -2(x - 2)2 - 5 e) y = 4(x - 5)2 - 2 f) y =
y
1 2 x - 2 3 v)
6 4 2 y
y
iii)
6 4 2
x -4 -2
0
x -2
0
-2 -4 -6
2
4
-2 -4 -6
2
4
6
8
-2 -4
2
4
6
ii)
6 4 2
y
iv)
6 4 2
y
vi)
-4 x -2
2
0
y x
-2 -4 -6 -8
2
4
6
x -4 -2
0
-2 -4 -6
2
4
6
-6 -4
-2
0
-2 -4 -6
2
4
-10
7. Sketch the graph of each quadratic relation by hand. Start with
a sketch of y = x 2, and then apply the appropriate transformations in the correct order. 3 a) y = -(x - 2)2 d) y = x 2 - 5 4 1 1 b) y = (x + 2)2 - 8 e) y = (x - 2)2 - 5 2 2 2 + 7 c) y = -3(x - 1) f ) y = -1.5(x + 3)2 + 10
270
5.3 Graphing Quadratics in Vertex Form Draft Evaluation Copy
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5.3
8. Copy and complete the following table. Stretch/ Compression Factor Horizontal/ Vertical Translation right 2, down 5
Quadratic Relation
Reflection in x-axis
Vertex
Axis of Symmetry
3 y = 4(x + 2)2 - 3 y = -(x - 1)2 + 4 y = 0.8(x - 6)2 y = 2x 2 - 5
no
(2,
5)
x
2
9. Determine the equations of three different parabolas with a vertex
C
at ( 2, 3). Describe how the graphs of the parabolas are different from each other. Then sketch the graphs of the three relations on the same set of axes. metres, after t seconds is modelled by h = -0.5(g - r)t 2 + k, where g is the acceleration due to gravity, r is the resistance offered by the parachute, and k is the height from which the object is dropped. On Earth, g 9.8 m/s 2. The resistance offered by a single bed sheet is 0.6 m/s 2, by a car tarp is 2.1 m/s 2, and by a regular parachute is 8.9 m/s 2. a) Describe how the graphs will differ for objects dropped from a height of 100 m using each of the three types of parachutes. b) Is it possible to drop an object attached to the bed sheet and a similar object attached to a regular parachute and have them hit the ground at the same time? Describe how it would be possible and what the graphs of each might look like.
10. When an object with a parachute is released to fall freely, its height, h, in
11. Write the equation of a parabola that matches each description.
a) The graph of y = x 2 is reflected about the x-axis and then translated 5 units up. b) The graph of y = x 2 is stretched vertically by a factor of 5 and then translated 2 units left. 1 c) The graph of y = x 2 is compressed vertically by a factor of 5 and then translated 6 units down. d) The graph of y = x 2 is reflected about the x-axis, stretched vertically by a factor of 6, translated 3 units right, and translated 4 units up.
12. Sketch the graph of each parabola described in question 11 by applying
K
the given sequence of transformations. Use a separate grid for each graph.
Draft Evaluation Copy Chapter 5
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271
6 4 2
y
13. Which equation represents the graph shown at the left? Explain your
reasoning.
x
-2
0
-2
2
4
6
8
2 c) y = - (x - 3)2 + 5 3 2 b) y = -(x - 3)2 + 5 d) y = (x - 3)2 + 5 3 14. A sky diver jumped from an airplane. He used his watch to time the length of his jump. His height above the ground can be modelled by h = -5(t - 4)2 + 2500, where h is his height above the ground in metres and t is the time in seconds from the time he started the timer. a) How long did the sky diver delay his jump? b) From what height did he jump? a) y = 15. A video tracking device recorded the height, h, in metres, of a baseball
A
2 2 x + 5 3
after it was hit. The data collected can be modelled by h = -5(t - 2)2 + 21, where t is the time in seconds after the ball was hit. a) Sketch a graph that represents the height of the baseball. b) What was the maximum height reached by the baseball? c) When did the baseball reach its maximum height? d) At what time(s) was the baseball at a height of 10 m? e) Approximately when did the baseball hit the ground? (8, 17). Describe three sets of transformations that could make this happen. For each set, give the equation of the new parabola. form and vertex form.
16. When a graph of y = x 2 is transformed, the point (3, 9) moves to
T
17. Express the quadratic relation y = 2(x - 4)(x + 10) in both standard 18. Copy and complete the chart to show what you know about the
quadratic relation y = -2(x + 3)2 + 4.
Translation: Reflection:
Stretch/ Compression:
y
2(x
3)2
4
Vertex:
Extending
19. Determine one of the zeros of the quadratic relation
y = ax 272
5.3 Graphing Quadratics in Vertex Form
k 2 (k - 2)2 . b 2 4
Draft Evaluation Copy
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Q:
A:
Mid-Chapter Review
FREQUENTLY ASKED Questions
How do you know whether the graph of y ax2 will have a wider or narrower shape near its vertex, compared with the graph of y x 2?
Study
Aid
• See Lesson 5.1,
Examples 1 and 2.
The shape depends on the value of a in the equation. Each y-value is multiplied by a factor of a. When a 1, the y-values increase. The parabola appears to be vertically stretched and becomes narrower near its vertex. When 0 a 1, the y-values decrease. The parabola appears to be vertically compressed and becomes wider near its vertex.
Why is the vertex form, y a(x h)2 for graphing quadratic relations? k, useful
• Try Mid-Chapter Review
Questions 1 and 2.
Q:
A1:
Study
Aid
You can use the constants a, h, and k to determine how the graph of y = x 2 has been transformed. • When a 1, the parabola is vertically stretched and when 0 a 1, the parabola is vertically compressed. • When a 0, the parabola is reflected in the x-axis. • The parabola is translated to the right when h 0 and to the left when h 0. The parabola is translated up when k 0 and down when k 0. • The coordinates of the vertex are (h, k). You can use the constants a, h, and k to determine key features of the parabola. • When a 0, the parabola opens upward. When a 0, the parabola opens downward. • The coordinates of the vertex are (h, k). • The equation of the axis of symmetry is x = h. You can use these properties, as well as the coordinates of a few other points, to draw an accurate sketch of any parabola.
• See Lesson 5.3, Examples 1 to 3. • Try Mid-Chapter Review Questions 3, 4, 6, and 7.
A2:
Q:
A:
When you use transformations to sketch a graph, why is the order in which you apply the transformations important?
Study
1 to 3.
Aid
When a graph is transformed, operations are performed on the coordinates of each point. Apply transformations in the same order you would apply calculations. Apply vertical stretches/compressions and reflections (multiplication) before translations (addition or subtraction).
Stretch Reflect Translate
Draft Evaluation Copy
• See Lesson 5.3, Examples • Try Mid-Chapter Review
Question 5.
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273
PRACTICE Questions
Lesson 5.1
4. The parabolas
1. Sketch the graph of each equation by correctly
applying the required transformation(s) to points on the graph of y = x 2. Use a separate grid for each graph. a) y = 2x 2 c) y = -3x 2 2 b) y = -0.25x 2 d) y = x 2 3 2. Describe the transformation(s) that were applied to the graph of y = x 2 to obtain each black graph. Write the equation of the black graph. a) y
8 y x2 6 4 2 x -4 -2
0
displayed on the graphing calculator screen were entered as equations of the form y = (x - h)2 + k. Determine as many of the equations as you can.
Lesson 5.3
5. Describe the sequence of transformations that
-2 6 4 2 y
2
4
you would apply to the graph of y = x 2 to sketch each quadratic relation. a) y = -3(x - 1)2 1 b) y = (x + 3)2 - 8 2 c) y = 4(x - 2)2 - 5 2 d) y = x 2 - 1 3
6. Sketch a graph of each quadratic relation in
b)
y
x2
question 5 on a separate grid. Use the properties of the parabola and some additional points.
7. For each quadratic relation,
x -4 -2
0
-2 -4 -6
2
4
i) state the stretch/compression factor and the horizontal/vertical translations ii) determine whether the graph is reflected in the x-axis iii) state the vertex and the equation of the axis of symmetry iv) sketch the graph by applying transformations to the graph of y = x 2 a) y = (x - 2)2 + 1 1 b) y = - (x + 4)2 2 c) y = 2(x + 1)2 - 8 d) y = -0.25x 2 + 5
8. A parabola lies in only two quadrants. What
Lesson 5.2
3. Determine the values of h and k for each of the
following transformations. Write the equation in the form y = (x - h)2 + k. Sketch the graph. a) The parabola moves 3 units down and 2 units right. b) The parabola moves 4 units left and 6 units up.
does this tell you about the values of a, h, and k? Explain your thinking, and provide the equation of a parabola as an example.
274
Mid-Chapter Review
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5.4
GOAL
Quadratic Models Using Vertex Form
YOU WILL NEED • grid paper • ruler • graphing calculator • spreadsheet program (optional) Bakery Profits from Bread Sales y (1.75, 400) 400 350 Profit ($) 300 250 200 (0.75, 300)
Write the equation of the graph of a quadratic relation in vertex form.
LEARN ABOUT the Math
The Best Bread Bakery wants to determine its daily profit from bread sales. This graph shows the data gathered by the company.
?
What equation represents the relationship between the price of bread and the daily profit from bread sales?
150 0 x 0.5 1 1.5 2 2.5 Price per loaf ($) 3
EXAMPLE
1
Connecting a parabola to the vertex form of its equation
Determine the equation of this quadratic relation from its graph. Sabrina’s Solution y = a(x - h)2 + k y = a(x - 1.75)2 + 400 300 300 300 –100 = = = = a(0.75 - 1.75)2 + 400 a(–1)2 + 400 a + 400 a
Since the graph is a parabola and the coordinates of the vertex are given, I decided to use vertex form. Since (1.75, 400) is the vertex, h 1.75 and k 400. I substituted these values into the equation. To determine the value of a, I chose the point (0.75, 300) on the graph. I substituted these coordinates for x and y in the equation. I followed the order of operations and solved for the value of a.
The equation that represents the relationship is y = –100(x - 1.75)2 + 400.
Reflecting
A.
What information do you need from the graph of a quadratic relation to determine the equation of the relation in vertex form? You have used the standard, factored, and vertex forms of a quadratic relation. Which form do you think is most useful for determining the equation of a parabola from its graph? Explain why.
Draft Evaluation Copy Chapter 5
B.
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APPLY the Math
EXAMPLE
2
Connecting information about a parabola to its equation
The graph of y = x 2 was stretched by a factor of 2 and reflected in the x-axis. The graph was then translated to a position where its vertex is not visible in the viewing window of a graphing calculator. Determine the quadratic relation in vertex form from the partial graph displayed in the screen shot. The scale factor on the y-axis is 5, and the scale factor on the x-axis is 2. Terri’s Solution a = -2 y = -2(x - h)2 + k
18 4
The graph was stretched by a factor of 2 and reflected in the x-axis. I substituted the value of a into the vertex form of the quadratic relation.
The zeros of the graph are 3 and 13. 3 + 13 h = 2 h = 8 18 18 18 50 = = = = -2(4 - 8)2 + k -2(16) + k -32 + k k
I determined the mean of the two zeros to calculate the value of h. The vertex lies on the axis of symmetry, which is halfway between the zeros of the graph. I saw that (4, 18) is a point on the graph. By substituting these coordinates, as well as the value I determined for h, I was able to solve for k.
The equation of the graph is y = -2(x - 8)2 + 50.
EXAMPLE
3
Selecting a strategy to determine a quadratic model
The amount of gasoline that a car consumes depends on its speed. A group of students decided to research the relationship between speed and fuel consumption for a particular car. They collected the data in the table. Determine an equation that models the relationship between speed and fuel consumption.
Speed (km/h) Gas Consumed (litres/100 km) 10 9.2 20 8.1 30 7.4 40 7.2 50 6.4 60 6.1 70 5.9 80 5.8 90 6.0 100 6.3 110 7.5 120 8.4
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Eric’s Solution: Representing a relation with a scatter plot and determining the equation algebraically
Gas (litres/100 km) 10 -8 6 4 2 0 vertex (75, 5.8) x 20 40 60 80 100 120 140 Speed (km/h) y I constructed a scatter plot to display the data and drew a curve of good fit. Since the curve looked parabolic and I knew that I could estimate the coordinates of the vertex. I estimated the coordinates of the vertex to be about (75, 5.8).
y = a(x - h)2 + k y = a(x - 75)2 + 5.8
I decided to use the vertex form of the equation. I substituted the estimated values (75, 5.8) into the general equation.
6.0 = a(90 - 75)2 + 5.8
From the table, I knew that the point (90, 6.0) is close to the curve. I substituted the coordinates of this point for x and y to determine a. I solved for a.
6.0 = a(15)2 + 5.8 6.0 = 225a + 5.8 0.2 = 225a 0.0009 a The equation that models the data is y = 0.0009(x - 75)2 + 5.8.
10.0 Gas (L/100 km) 8.0 6.0 4.0 2.0 0.0 0 50 100 Speed (km/h) 150 y 0.0009x2 10.534 0.1317x Gas Consumption
I checked my equation using a spreadsheet. I entered the data from the table. I used column A for the Speed values and column B for the Gas values. I created a graph, added a trend line using quadratic regression of order 2, and chose the option to display the equation on the graph. Tech
Support
For help creating a scatter plot and performing a regression analysis using a spreadsheet, see Appendix B-35.
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y y y y
= = = =
0.0009(x - 75)2 + 5.8 0.0009(x 2 - 150x + 5625) + 5.8 0.0009x 2 - 0.135x + 5.0625 + 5.8 0.0009x 2 - 0.135x + 10.8625
The spreadsheet equation was in standard form, but my equation was in vertex form. To compare the two equations, I expanded my equation.
The two equations are very close, so they are both good quadratic models for this set of data. Gillian’s Solution: Selecting a graphing calculator and an informal curve-fitting process
I entered the data into L1 and L2 in the data editor of a graphing calculator and created a scatter plot.
y = a(x - 75)2 + 5.8
The points had a parabolic pattern, so I estimated the coordinates of the vertex to be about (75, 5.8). I substituted these coordinates into the general equation.
Since the parabola opens upward, I knew that a 7 0. I used a = 1 and entered the equation y = 1(x - 75)2 + 5.8 into Y1 of the equation editor. Then I graphed the equation. The location of the vertex looked good, but the parabola wasn’t wide enough.
I decreased the value of a to a = 0.1, but the parabola still wasn’t wide enough.
I decreased the value of a several more times until I got a good fit. I found that a = 0.0009 worked fairly well. Tech
An equation that models the relationship between speed and fuel consumption is y = 0.0009(x - 75)2 + 5.8.
Support
For help creating a scatter plot using a graphing calculator, see Appendix B-10.
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I checked my equation by comparing it with the equation produced by quadratic regression on the graphing calculator. To do this, I had to expand my equation.
y y y y
= = = =
0.0009(x - 75)2 + 5.8 0.0009(x 2 - 150x + 5625) + 5.8 0.0009x 2 - 0.135x + 5.0625 + 5.8 0.0009x 2 - 0.135x + 10.8625
Tech
Support
The two equations are close, so they are both good models for this set of data.
For help performing a quadratic regression analysis using a graphing calculator, see Appendix B-10.
In Summary
Key Idea
• If you know the coordinates of the vertex (h, k) and one other point on a parabola, you can determine the equation of the relation using y = a(x - h)2 + k.
Need to Know
• To determine the value of a, substitute the coordinates of a point on the graph into the general equation and solve for a: • If (h, k) = (., .), then y = a(x - .)2 + .. • If a point on the graph has coordinates x = . and y = ., then, by substitution, . = a(. - .)2 + .. • Since a is the only remaining unknown, its value can be determined by solving the equation. • The vertex form of an equation y (h, k) can be determined using the zeros of the graph. The axis of symmetry is x = h, where h is the mean of the zeros. • You can convert a quadratic x (r, 0) (s, 0) equation from vertex form to standard form by expanding and then collecting like terms. r s x — h 2
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CHECK Your Understanding
1. Match each equation with the correct graph.
a) y = 2x 2 – 8 b) y = (x + 3)2 i)
10 8 6 4 2 y
c) y = -2(x – 3)2 + 8 d) y = (x – 3)2 – 8 iii)
4 2 x -4 -2
0
y
-2 -4
2
4
x -2
0
-6 -8 -10
-2 -4
2
4
6
8
ii)
4 2
y
iv)
x
10 8
y
-2
0
6 4 2 x -8 -6 -4 -2
0
-2 -4 -6 -8
2
4
6
8
-2 -4
2
-10
2. The vertex of a quadratic relation is (4,
12). a) Write an equation to describe all parabolas with this vertex. b) A parabola with the given vertex passes through point (13, 15). Determine the value of a for this parabola. c) Write the equation of the relation in part b). d) State the transformations that must be applied to y x2 to obtain the quadratic relation you wrote for part c). e) Graph the quadratic relation you wrote for part c).
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PRACTISING
3. Write the equation of each parabola in vertex form.
a)
6 4 2
y
c)
6 4 2
y
e)
6 4 2
y
(2, 1) x -4 -2
0
(2, 0) -4 -2
0
x -4 -2
0
(3, 1) 2 4
x
-2 y
2
4
-2 y
2
4
-2 y
b)
6 4 (–2, 2) -4 -2 2
d)
4 2
f)
x (–3, 2) -4 -2
4 2
x
0
x
0
-2
2
4
-4 -2 (–1, –2) -4
0
2
4
-2 -4
2
4
4. The following transformations are applied to the graph of y = x 2.
Determine the equation of each new relation. a) a vertical stretch by a factor of 4 b) a translation of 3 units left c) a reflection in the x-axis, followed by a translation 2 units up d) a vertical compression by a factor of 1 2 e) a translation of 5 units right and 4 units down f ) a vertical stretch by a factor of 2, followed by a reflection in the x-axis and a translation 1 unit left
5. Write the equation of a parabola with each set of properties.
vertex at (0, 4), opens upward, the same shape as y x2 vertex at (5, 0), opens downward, the same shape as y x2 vertex at (2, 3), opens upward, narrower than y x2 vertex at ( 3, 5), opens downward, wider than y x2 axis of symmetry x 4, opens upward, two zeros, narrower than y x2 f ) vertex at (3, – 4), no zeros, wider than y x2 a) b) c) d) e)
6. Determine the equation of a quadratic relation in vertex form, given
K
the following information. a) vertex at ( 2, 3), passes through ( 4, 1) b) vertex at ( 1, 1), passes through (0, 1) c) vertex at ( 2, 3), passes through ( 5, 6) d) vertex at ( 2, 5), passes through (1, 4)
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7. Each table of values defines a parabola. Determine the axis of
symmetry of the parabola, and write the equation in vertex form. a)
x 2 3 4 5 6 y 33 13 1 3 1
b)
x 0 1 2 3 4
y 12 4 4 12 28
8. A child kicks a soccer ball so that it barely clears a 2 m fence. The
soccer ball lands 3 m from the fence. Determine the equation, in vertex form, of a quadratic relation that models the path of the ball.
9. Data for DVD sales in Canada, over several years, are given in the table. Year x, Years Since 2002 DVDs Sold (1000s) 2002 0 1446 2003 1 3697 2004 2 4573 2005 3 4228 2006 4 3702
a) Using graphing technology, create a scatter plot to display the data. b) Estimate the vertex of the graph you created for part a). Then determine an equation in vertex form to model the data. c) How many DVDs would you expect to be sold in 2010? d) Check the accuracy of your model using quadratic regression.
10. A school custodian finds a tennis ball on the roof of the school and
A
throws it to the ground below. The table gives the height of the ball above the ground as it moves through the air.
Time (s) 0.0 0.5 11.25 1.0 15.00 1.5 16.25 2.0 15.00 2.5 11.25 3.0 5.00
Height (m) 5.00
a) b) c) d) e) f)
Do the data appear to be linear or quadratic? Explain. Create a scatter plot, and draw a quadratic curve of good fit. Estimate the coordinates of the vertex. Determine an algebraic relation in vertex form to model the data. Use your model to predict the height of the ball at 2.75 s and 1.25 s. How effective is your model for time values that are greater than 3.5 s? Explain. g) Check the accuracy of your model using quadratic regression.
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11. A chain of ice cream stores sells $840 of ice cream cones per day. Each
ice cream cone costs $3.50. Market research shows the following trend in revenue as the price of an ice cream cone is reduced.
Price ($) Revenue ($) 3.50 840 3.00 2520 2.50 3600 2.00 4080 1.50 3960 1.00 3240 0.50 1920
a) Create a scatter plot, and draw a quadratic curve of good fit. b) Determine an equation in vertex form to model this relation. c) Use your model to predict the revenue if the price of an ice cream cone is reduced to $2.25. d) To maximize revenue, what should an ice cream cone cost? e) Check the accuracy of your model using quadratic regression.
12. This table shows the number of imported cars that were sold
in Newfoundland between 2003 and 2007.
Year Sales of Imported Cars (number sold) 2003 3996 2004 3906 2005 3762 2006 3788 2007 4151
a) Create a scatter plot, and draw a quadratic curve of good fit. b) Determine an algebraic equation in vertex form to model this relation. c) Use your model to predict how many imported cars were sold in 2008. d) What does your model predict for 2006? Is this prediction accurate? Explain. e) Check the accuracy of your model using quadratic regression.
13. The Lion’s Gate Bridge in Vancouver, British Columbia, is a
T
suspension bridge that spans a distance of 1516 m. Large cables are attached to the tops of the towers, 50 m above the road. The road is suspended from the large cables by many smaller vertical cables. The smallest vertical cable measures about 2 m. Use this information to determine a quadratic model for the large cables. reaches a maximum height of 2000 m. It lands on the ground after 40 s. Explain how you could determine the equation of the relationship between the height of the rocket and time using two different strategies.
14. A model rocket is launched from the ground. After 20 s, the rocket
C
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15. The owner of a small clothing company wants to create a
mathematical model for the company’s daily profit, p, in dollars, based on the selling price, d, of the dresses made. The owner has noticed that the maximum daily profit the company has made is $1600. This occurred when the dresses were sold for $75 each. The owner also noticed that selling the dresses for $50 resulted in a profit of $1225. Using a quadratic relation to model this problem, create an equation for the company’s daily profit.
16. Compare the three forms of the equation of a quadratic relation using
this concept circle. Under what conditions would you use one form instead of the other forms when trying to connect a graph to its equation? Explain your thinking.
Quadratic Relations
Factored Form
Standard Form
Vertex Form
Extending
17. The following transformations are applied to a parabola with the
equation y = 2(x + 3)2 – 1. Determine the equation that will result after each transformation. a) a translation 4 units right b) a reflection in the x-axis c) a reflection in the x-axis, followed by a translation 5 units down d) a stretch by a factor of 6 e) a compression by a factor of 1, followed by a reflection in the y-axis 4
18. The vertex of the parabola y = 3x 2 + bx + c is at ( 1, 4).
Determine the values of b and c.
19. Determine an algebraic expression for the solution, x, to the equation
0 = a(x - h)2 + k. Do not expand the equation.
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GOAL
Solving Problems Using Quadratic Relations
YOU WILL NEED • grid paper • ruler
Model and solve problems using the vertex form of a quadratic relation.
LEARN ABOUT the Math
Smoke jumpers are firefighters who parachute into remote locations to suppress forest fires. They are often the first people to arrive at a fire. When smoke jumpers exit an airplane, they are in free fall until their parachutes open. A quadratic relation can be used to determine the height, H, in metres, of a jumper t seconds after exiting an airplane. In this relation, a = -0.5g, where g is the acceleration due to gravity. On Earth, g = 9.8 m/s2.
?
If a jumper exits an airplane at a height of 554 m, how long will the jumper be in free fall before the parachute opens at 300 m?
EXAMPLE
1
Connecting information from a problem to a quadratic model
a) Determine the quadratic relation that will model the height, H, of the smoke jumper at time t. b) Determine the length of time that the jumper is in free fall. Conor’s Solution
I decided to use the vertex form of the quadratic relation because the problem contains information about the vertex.
a) H = a(t - h)2 + k
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H = a(t - 0)2 + 554
The vertex is the point at which the jumper exited the plane. So the vertex has coordinates (0, 554). I substituted these coordinates into the general equation. Since a 0.5g and g 9.8 m/s2, I substituted these values into the vertex form of the equation.
H = -0.5(9.8)(t - 0)2 + 554 H = -4.9(t - 0)2 + 554 is an equation in vertex form for the quadratic relation that models this situation. H = -4.9t 2 + 554 is an equation in standard form for the quadratic relation that models this situation. b) 300 -254 -254 -4.9 51.84 151.84 7.2 = -4.9t 2 + 554 = -4.9t 2 -4.9t 2 = -4.9 2 = t = t t, since t 7 0
I noticed that the value of a is the same in both vertex form and standard form. This makes sense because the parabolas would not be congruent if they were different. Because the parachute opened at 300 m, I substituted 300 for H. Then I solved for t.
The jumper is in free fall for about 7.2 s.
In this situation, time can’t be negative. So, I didn’t use the negative square root of 51.84.
Reflecting
A. B.
Why was zero used for the t-coordinate of the vertex? How would the equation change if the jumper hesitated for 2 s before exiting the airplane, after being given the command to jump? Why was the vertex form easier to use than either of the other two forms of a quadratic relation in this problem?
C.
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APPLY the Math
EXAMPLE
2
Solving a problem using a quadratic model
The underside of a concrete railway underpass forms a parabolic arch. The arch is 30.0 m wide at the base and 10.8 m high in the centre. The upper surface of the underpass is 40.0 m wide. The concrete is 2.0 m thick at the centre. Can a truck that is 5 m wide and 7.5 m tall get through this underpass? Lisa’s Solution
y 4 2 (0, 0) -20 -16 -12 -8 -4
0
x 8 12 16 20
(–15, –10.8)
-4 -6 -8 -10 -12 -14
4
(15, –10.8)
I started by drawing a diagram. I used a grid and marked the top of the arch as (0, 0). The upper surface of the underpass is 2 m above the top of the arch at the centre. The arch is 10.8 m high in the centre and 30 m wide at the base (or 15 m wide on each side). I marked the points ( 15, 10.8) and (15, 10.8) and drew a parabola through these two points and the origin. The vertex of the parabola is at the origin, so I did not translate y = ax 2. I determined the value of a by substituting the coordinates of a point on the graph, (15, 10.8), into this equation. Then I solved for a.
y = ax 2 -10.8 = a(15)2 -10.8 = 225a -0.048 = a y = -0.048x 2 is the quadratic relation that models the arch of the railway underpass.
y 4 2 (0, 0) -20 -16 -12 -8 -4
0
x 8 12 16 20
(–15, –10.8)
-4 -6 -8 -10 -12 -14
4
(15, –10.8)
The truck has the best chance of getting through the underpass if it passes through the centre. Since the truck is 5 m wide, this means that the position of the right corner of the truck has an x-coordinate of 2.5. I substituted x = 2.5 into the equation to check the height of the underpass at this point.
y = -0.048(2.5)2 y = -0.048(6.25) y = -0.3 Height at (2.5, -0.3) = 10.8 - 0.3 = 10.5 The truck can get through. Since the truck is 7.5 m tall, there is 3 m of clearance.
NEL
I determined the height from the ground at this point by subtracting 0.3 from 10.8. The truck can get through the overpass, even if it is a little off the centre of the overpass.
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EXAMPLE
3
Selecting a strategy to determine the vertex form
Write the quadratic relation y = x 2 - 4x - 5 in vertex form, and sketch the graph by hand. Coral’s Solution y = x 2 - 4x - 5 y = (x + 1)(x - 5)
I rewrote the equation of the quadratic relation in factored form because I knew that I could determine the coordinates of the vertex from this form. I set y 0 to determine the zeros. I used the zeros to determine the equation of the axis of symmetry.
Zeros: 0 = (x + 1)(x - 5) x = -1 and x = 5 The axis of symmetry is -1 + 5 so x 2. x = 2 y = (2)2 - 4(2) - 5 y = 4 - 8 - 5 y = -9
I substituted x = 2 into the standard form of the equation to solve for y.
The vertex is at (h, k) = (2, -9). The coefficient of x 2 is a = 1.
I knew that the value of a must be the same in the standard, factored, and vertex forms. If it were different, the parabola would have different widths. I substituted what I knew into the vertex form, y = a (x - h)2 + k.
The relation is y = (x - 2)2 - 9 in vertex form.
16 y x2 12 8 4 x -6 -4 -2
0
y
-4 -8 y
2
4
6
8
I sketched the graph of y = x 2 and translated each point 2 units right and 9 units down.
-12 -16
(x
2)2
9
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EXAMPLE
4
Representing a situation with a quadratic model
The Next Cup coffee shop sells a special blend of coffee for $2.60 per mug. The shop sells about 200 mugs per day. Customer surveys show that for every $0.05 decrease in the price, the shop will sell 10 more mugs per day. a) Determine the maximum daily revenue from coffee sales and the price per mug for this revenue. b) Write an equation in both standard form and vertex form to model this problem. Then sketch the graph. Dave’s Solution: Connecting the zeros of a parabola to the vertex form of the equation a) Let x represent the number of $0.05 decreases in price, where Revenue (price)(mugs sold).
I defined a variable that connects the price per mug to the number of mugs sold. I used the information in the problem to write expressions for the price per mug and the number of mugs sold in terms of x. If I drop the price by $0.05, x times, then the price per mug is 2.60 0.05x and the number of mugs sold is 200 10x. I used my expressions to write a relationship for daily revenue, r.
r = (2.60 - 0.05x)(200 + 10x)
0 = (2.60 - 0.05x)(200 + 10x) 2.60 - 0.05x = 0 or 200 + 10x = 0 x = 52 and x = -20 52 + (-20) 2 x = 16 x =
Since the equation is in factored form, the zeros of the equation can be calculated by letting r 0 and solving for x.
I used the zeros to determine the equation of the axis of symmetry. The maximum value occurs at the vertex. To calculate it, I substituted, the x-value for the axis of symmetry into the revenue equation.
r = [2.60 - 0.05(16)][200 + 10(16)] r = (1.80)(360) r = 648 The maximum daily revenue is $648.
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Price per mug for maximum revenue = 2.60 - 0.05(16) = 1.80 The coffee shop should sell each mug of coffee for $1.80 to achieve a maximum daily revenue of $648. b) In this relation, the maximum value is r = 648. It occurs when x = 16. The vertex is (16, 648) (h, k). r = a(x - 16)2 + 648 When x = 0, r = (2.60)(200) = 520 520 = a(0 - 16)2 + 648 520 = a(-16)2 + 648 -128 = 256a -0.5 = a The equation in vertex form is r = -0.5(x - 16)2 + 648.
I substituted x 16 into the expression for the price per mug.
The maximum value occurs at the vertex of a quadratic relation. To write the equation in vertex form, substitute the values of h and k into the vertex form of the general equation. Since the coffee shop sells 200 mugs of coffee when the price is $2.60 per mug, the point (0, 520) is on the graph. I substituted these coordinates into the equation and solved for a.
r r r r
= = = =
-0.5(x - 16)2 + 648 -0.5(x 2 - 32x + 256) + 648 -0.5x 2 + 16x - 128 + 648 -0.5x 2 + 16x + 520
I expanded to get the equation in standard form.
The equation in standard form is r = -0.5x 2 + 16x + 520.
700 600 500 400 300 200 100 x -20 -10
0
r
(16, 648) The vertex is at (16, 648). The zeros are at (52, 0) and ( 20, 0). The y-intercept is at (0, 520). I used these points to sketch the graph of the relation.
10 20 30 40 50
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