Main menu
This is “Graphing Parabolas”, section 9.5 from the book Beginning Algebra (v. 1.0). For details on it (including licensing), click here.
For more information on the source of this book, or why it is available for free, please see the project's home page. You can browse or download additional books there.
Has this book helped you? Consider passing it on:
Help Creative Commons
Creative Commons supports free culture from music to education. Their licenses helped make this book available to you.
Help a Public School
DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.
Previous Section
Table of Contents
Next Section
9.5 Graphing Parabolas
Learning Objectives
1.Graph a parabola.
2.Find the intercepts and vertex of a parabola.
3.Find the vertex of a parabola by completing the square.
The Graph of a Quadratic Equation
We know that any linear equation with two variables can be written in the form y=mx+b and that its graph is a line. In this section, we will see that any quadratic equation of the form y=ax2+bx+c has a curved graph called a parabolaThe graph of any quadratic equation y=ax2+bx+c, where a, b, and c are real numbers and a≠0..
Two points determine any line. However, since a parabola is curved, we should find more than two points. In this text, we will determine at least five points as a means to produce an acceptable sketch. To begin, we graph our first parabola by plotting points. Given a quadratic equation of the form y=ax2+bx+c, x is the independent variable and y is the dependent variable. Choose some values for x and then determine the corresponding y-
Example 1: Graph by plotting points: y=x2−2x−3.
Solution: In this example, choose the x-
Plot these points and determine the shape of the graph.
Answer:
When graphing, we want to include certain special points in the graph. The y-
For any parabola, we will find the vertex and y-
Given a quadratic equation of the form y=ax2+bx+c, find the y-
Next, recall that the x-
Using the fact that a parabola is symmetric, we can determine the vertical line of symmetry using the x-
Therefore, the line of symmetry is the vertical line:
We can use the line of symmetry to find the x-
Example 2: Graph: y=−x2−2x+3.
Solution:
Step 1: Determine the y-
The y-
Step 2: Determine the x-
Here when y = 0, we obtain two solutions. There are two x-
Step 3: Determine the vertex. One way to do this is to use the equation for the line of symmetry, x=−b2a, to find the x-
Substitute −1 into the original equation to find the corresponding y-
The vertex is (−1, 4).
Step 4: Determine extra points so that we have at least five points to plot. In this example, one other point will suffice. Choose x = −2 and find the corresponding y-
Our fifth point is (−2, 3).
Step 5: Plot the points and sketch the graph. To recap, the points that we have found are
y-
x-
Vertex: (−1, 4)
Extra point: (−2, 3)
Answer:
The parabola opens downward. In general, use the leading coefficient to determine whether the parabola opens upward or downward. If the leading coefficient is negative, as in the previous example, then the parabola opens downward. If the leading coefficient is positive, then the parabola opens upward.
All quadratic equations of the form y=ax2+bx+c have parabolic graphs with y-
Example 3: Graph: y=2x2+4x+5.
Solution: Because the leading coefficient 2 is positive, note that the parabola opens upward. Here c = 5 and the y-
In this case, a = 2, b = 4, and c = 5. Use the discriminant to determine the number and type of solutions.
Since the discriminant is negative, we conclude that there are no real solutions. Because there are no real solutions, there are no x-
Given that the x-
The vertex is (−1, 3). So far, we have only two points. To determine three more, choose some x-
To summarize, we have
y-
x-
Vertex: (−1, 3)
Extra points: (−3, 11), (−2, 5), (1, 11)
Plot the points and sketch the graph.
Answer:
Example 4: Graph: y=−2x2+12x−18.
Solution: Note that a = −2: the parabola opens downward. Since c = −18, the y-
Solve by factoring.
Here x = 3 is a double root, so there is only one x-
Given that the x-
Therefore, the vertex is (3, 0), which happens to be the same point as the x-
To summarize, we have
y-
x-
Vertex: (3, 0)
Extra points: (1, −8), (5, −8), (6, −18)
Plot the points and sketch the graph.
Answer:
Example 5: Graph: y=x2−2x−1.
Solution: Since a = 1, the parabola opens upward. Furthermore, c = −1, so the y-
In this case, solve using the quadratic formula with a = 1, b = −2, and c = −1.
Here we obtain two real solutions for x, and thus there are two x-
Approximate values using a calculator:
Use the approximate answers to place the ordered pair on the graph. However, we will present the exact x-
Given that the x-
The vertex is (1, −2). We need one more point.
To summarize, we have
y-
x-
Vertex: (1, −2)
Extra point: (2, −1)
Plot the points and sketch the graph.
Answer:
Try this! Graph: y=9x2−5.
Answer:
Video Solution
(click to see video)
Finding the Maximum and Minimum
It is often useful to find the maximum and/or minimum values of functions that model real-
Example 6: Determine the maximum or minimum: y=−4x2+24x−35.
Solution: Since a = −4, we know that the parabola opens downward and there will be a maximum y-
The x-
The vertex is (3, 1). Therefore, the maximum y-
Note
The graph is not required to answer this question.
Answer: The maximum is 1.
Example 7: Determine the maximum or minimum: y=4x2−32x+62.
Solution: Since a = +4, the parabola opens upward and there is a minimum y-
Substitute x = 4 into the original equation to find the corresponding y-
The vertex is (4, −2). Therefore, the minimum y-
Answer: The minimum is −2.
Try this! Determine the maximum or minimum: y=(x−3)2−9.
Answer: The minimum is −9.
Video Solution
(click to see video)
A parabola, opening upward or downward (as opposed to sideways), defines a function and extends indefinitely to the right and left as indicated by the arrows. Therefore, the domain (the set of x-
Example 8: Determine the domain and range: y=x2−4x+3.
Solution: First, note that since a=1 is positive, the parabola opens upward. Hence there will be a minimum y-
Then substitute into the equation to find the corresponding y-
The vertex is (2, −1). The range consists of the set of y-
Answer: Domain: R = (−∞, ∞); range: [−1, ∞)
Example 9: The height in feet of a projectile is given by the function h(t)=−16t2+72t, where t represents the time in seconds after launch. What is the maximum height reached by the projectile?
Solution: Here a=−16, and the parabola opens downward. Therefore, the y-
The maximum height will occur in 9/4 = 2¼ seconds. Substitute this time into the function to determine the height attained.
Answer: The maximum height of the projectile is 81 feet.
Finding the Vertex by Completing the Square
In this section, we demonstrate an alternate approach for finding the vertex. Any quadratic equation y=ax2+bx+c can be rewritten in the form
In this form, the vertex is (h, k).
Example 10: Determine the vertex: y=−4(x−3)2+1.
Solution: When the equation is in this form, we can read the vertex directly from the equation.
Here h = 3 and k = 1.
Answer: The vertex is (3, 1).
Example 11: Determine the vertex: y=2(x+3)2−2.
Solution: Rewrite the equation as follows before determining h and k.
Here h = −3 and k = −2.
Answer: The vertex is (−3, −2).
Often the equation is not given in this form. To obtain this form, complete the square.
Example 12: Rewrite in y=a(x−h)2+k form and determine the vertex: y=x2+4x+9.
Solution: Begin by making room for the constant term that completes the square.
The idea is to add and subtract the value that completes the square, (b2)2, and then factor. In this case, add and subtract (42)2=(2)2=4.
Adding and subtracting the same value within an expression does not change it. Doing so is equivalent to adding 0. Once the equation is in this form, we can easily determine the vertex.
Here h = −2 and k = 5.
Answer: The vertex is (−2, 5).
If there is a leading coefficient other than 1, then we must first factor out the leading coefficient from the first two terms of the trinomial.
Example 13: Rewrite in y=a(x−h)2+k form and determine the vertex: y=2x2−4x+8.
Solution: Since a = 2, factor this out of the first two terms in order to complete the square. Leave room inside the parentheses to add a constant term.
Now use −2 to determine the value that completes the square. In this case, ( −22)2=(−1)2=1. Add and subtract 1 and factor as follows:
In this form, we can easily determine the vertex.
Here h = 1 and k = 6.
Answer: The vertex is (1, 6).
Try this! Rewrite in y=a(x−h)2+k form and determine the vertex: y=−2x2−12x+3.
Answer: y=−2(x+3)2+21; vertex: (−3, 21)
Video Solution
(click to see video)
Key Takeaways
•The graph of any quadratic equation y=ax2+bx+c, where a, b, and c are real numbers and a≠0, is called a parabola.
•When graphing parabolas, find the vertex and y-
•Use the leading coefficient, a, to determine if a parabola opens upward or downward. If a is positive, then it opens upward. If a is negative, then it opens downward.
•The vertex of any parabola has an x-
•The domain of a parabola opening upward or downward consists of all real numbers. The range is bounded by the y-
•An alternate approach to finding the vertex is to rewrite the quadratic equation in the form y=a(x−h)2+k. When in this form, the vertex is (h, k) and can be read directly from the equation. To obtain this form, take y=ax2+bx+c and complete the square.
Topic Exercises
Part A: The Graph of Quadratic Equations
Does the parabola open upward or downward? Explain.
1. y=x2−9x+20
2. y=x2−12x+32
3. y=−2x2+5x+12
4. y=−6x2+13x−6
5. y=64−x2
6. y=−3x+9x2
Determine the x-
7. y=x2+4x−12
8. y=x2−13x+12
9. y=2x2+5x−3
10. y=3x2−4x−4
11. y=−5x2−3x+2
12. y=−6x2+11x−4
13. y=4x2−25
14. y=9x2−49
15. y=x2−x+1
16. y=5x2+15x
Find the vertex and the line of symmetry.
17. y=−x2+10x−34
18. y=−x2−6x+1
19. y=−4x2+12x−7
20. y=−9x2+6x+2
21. y=4x2−1
22. y=x2−16
Graph. Find the vertex and the y-
23. y=x2−2x−8
24. y=x2−4x−5
25. y=−x2+4x+12
26. y=−x2−2x+15
27. y=x2−10x
28. y=x2+8x
29. y=x2−9
30. y=x2−25
31. y=1−x2
32. y=4−x2
33. y=x2−2x+1
34. y=x2+4x+4
35. y=−4x2+12x−9
36. y=−4x2−4x+3
37. y=x2−2
38. y=x2−3
39. y=−4x2+4x−3
40. y=4x2+4x+3
41. y=x2−2x−2
42. y=x2−6x+6
43. y=−2x2+6x−3
44. y=−4x2+4x+1
45. y=x2+3x+4
46. y=−x2+3x−4
47. y=−2x2+3
48. y=−2x2−1
49. y=2x2+4x−3
50. y=3x2+2x−2
Part B: Maximum or Minimum
Determine the maximum or minimum y-
51. y=−x2−6x+1
52. y=−x2−4x+8
53. y=25x2−10x+5
54. y=16x2−24x+7
55. y=−x2
56. y=1−9x2
57. y=20x−10x2
58. y=12x+4x2
59. y=3x2−4x−2
60. y=6x2−8x+5
Given the following quadratic functions, determine the domain and range.
61. f(x)=3x2+30x+50
62. f(x)=5x2−10x+1
63. g(x)=−2x2+4x+1
64. g(x)=−7x2−14x−9
65. The height in feet reached by a baseball tossed upward at a speed of 48 feet/second from the ground is given by the function h(t)=−16t2+48t, where t represents time in seconds. What is the baseball’s maximum height and how long will it take to attain that height?
66. The height of a projectile launched straight up from a mound is given by the function h(t)=−16t2+96t+4, where t represents seconds after launch. What is the maximum height?
67. The profit in dollars generated by producing and selling x custom lamps is given by the function P(x)=−10x2+800x−12,000. What is the maximum profit?
68. The revenue in dollars generated from selling a particular item is modeled by the formula R(x)=100x−0.0025x2, where x represents the number of units sold. What number of units must be sold to maximize revenue?
69. The average number of hits to a radio station website is modeled by the formula f(x)=450t2−3,600t+8,000, where t represents the number of hours since 8:00 a.m. At what hour of the day is the number of hits to the website at a minimum?
70. The value in dollars of a new car is modeled by the formula V(t)=125t2−3,000t+22,000, where t represents the number of years since it was purchased. Determine the minimum value of the car.
71. The daily production costs in dollars of a textile manufacturing company producing custom uniforms is modeled by the formula C(x)=0.02x2−20x+10,000, where x represents the number of uniforms produced.
a. How many uniforms should be produced to minimize the daily production costs?
b. What is the minimum daily production cost?
72. The area of a certain rectangular pen is given by the formula A=14w−w2, where w represents the width in feet. Determine the width that produces the maximum area.
Part C: Vertex by Completing the Square
Determine the vertex.
73. y=−(x−5)2+3
74. y=−2(x−1)2+7
75. y=5(x+1)2+6
76. y=3(x+4)2+10
77. y=−5(x+8)2−1
78. y=(x+2)2−5
Rewrite in y=a(x−h)2+k form and determine the vertex.
79. y=x2−14x+24
80. y=x2−12x+40
81. y=x2+4x−12
82. y=x2+6x−1
83. y=2x2−12x−3
84. y=3x2−6x+5
85. y=−x2+16x+17
86. y=−x2+10x
Graph.
87. y=x2−1
88. y=x2+1
89. y=(x−1)2
90. y=(x+1)2
91. y=(x−4)2−9
92. y=(x−1)2−4
93. y=−2(x+1)2+8
94. y=−3(x+2)2+12
95. y=−5(x−1)2
96. y=−(x+2)2
97. y=−4(x−1)2−2
98. y=9(x+1)2+2
99. y=(x+5)2−15
100. y=2(x−5)2−3
101. y=−2(x−4)2+22
102. y=2(x+3)2−13
Part D: Discussion Board
103. Write down your plan for graphing a parabola on an exam. What will you be looking for and how will you present your answer? Share your plan on the discussion board.
104. Why is any parabola that opens upward or downward a function? Explain to a classmate how to determine the domain and range.
Answers
1: Upward
3: Downward
5: Downward
7: x-
9: x-
11: x-
13: x-
15: x-
17: Vertex: (5, −9); line of symmetry: x=5
19: Vertex: (3/2, 2); line of symmetry: x=3/2
21: Vertex: (0, −1); line of symmetry: x=0
23:
25:
27:
29:
31:
33:
35:
37:
39:
41:
43:
45:
47:
49:
51: Maximum: y = 10
53: Minimum: y = 4
55: Maximum: y = 0
57: Maximum: y = 10
59: Minimum: y = −10/3
61: Domain: R; range: [−25,∞)
63: Domain: R; range: (−∞,3]
65: The maximum height of 36 feet occurs after 1.5 seconds.
67: $4,000
69: 12:00 p.m.
71: a. 500 uniforms; b. $5,000
73: (5, 3)
75: (−1, 6)
77: (−8, −1)
79: y=(x−7)2−25; vertex: (7, −25)
81: y=(x+2)2−16; vertex: (−2, −16)
83: y=2(x−3)2−21; vertex: (3, −21)
85: y=−(x−8)2+81; vertex: (8, 81)
87:
89:
91:
93:
95:
97:
99:
101:
Previous Section
Table of Contents
Next Section