How Do You Know Where a Parabola Opens

The Graph of a Quadratic Equation

We know that any linear equation with 2 variables can be written in the grade $y=\mathrm{one\; thousand}x+b$ and that its graph is a line. In this section, we volition run into that any quadratic equation of the class $y=a{x}^2+bx+c$ has a curved graph called a parabolaThe graph of whatever quadratic equation $y=a{\mathrm{10}}^2+bx+c$ , where a, b, and c are real numbers and $a\ne 0$ . .

Two points determine whatever line. Withal, since a parabola is curved, nosotros should discover more than than two points. In this text, we will determine at least five points as a means to produce an acceptable sketch. To begin, nosotros graph our starting time parabola by plotting points. Given a quadratic equation of the grade $y=a{\mathrm{ten}}^2+b\mathrm{ten}+c$ , x is the independent variable and y is the dependent variable. Choose some values for 10 and then determine the corresponding y-values. Then plot the points and sketch the graph.

Example 1: Graph by plotting points: $y={\mathrm{ten}}^2-2x-3$ .

Solution: In this example, choose the 10-values {−2, −1, 0, 1, 2, 3, iv} and calculate the corresponding y-values.

Plot these points and determine the shape of the graph.

Reply:

When graphing, we want to include certain special points in the graph. The y-intercept is the point where the graph intersects the y-axis. The x-intercepts are the points where the graph intersects the x-axis. The vertexThe point that defines the minimum or maximum of a parabola. is the point that defines the minimum or maximum of the graph. Lastly, the line of symmetryThe vertical line through the vertex, $\mathrm{ten}=-\frac{b}{2a}$ , most which the parabola is symmetric. (also called the axis of symmetryA term used when referencing the line of symmetry. ) is the vertical line through the vertex, about which the parabola is symmetric.

For any parabola, nosotros will observe the vertex and y-intercept. In addition, if the x-intercepts exist, then we will desire to make up one's mind those likewise. Guessing at the ten-values of these special points is not applied; therefore, nosotros will develop techniques that volition facilitate finding them. Many of these techniques will exist used extensively every bit we progress in our report of algebra.

Given a quadratic equation of the form $y=a{x}^2+bx+c$ , find the y-intercept by setting $\mathrm{10}=0$ and solving. In general, $y=a{\left(0\right)}^{\mathrm{ii}}+b\left(0\right)+c=c$ , and nosotros have

Next, recall that the x-intercepts, if they exist, can be found by setting $y=0$ . Doing this, nosotros have $0=a^2+bx+c$ , which has full general solutions given by the quadratic formula, $\mathrm{ten}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . Therefore, the 10-intercepts have this general class:

Using the fact that a parabola is symmetric, we can determine the vertical line of symmetry using the 10-intercepts. To practice this, we find the x-value midway betwixt the x-intercepts by taking an boilerplate as follows:

Therefore, the line of symmetry is the vertical line:

We can employ the line of symmetry to find the x-value of the vertex. The steps for graphing a parabola are outlined in the post-obit example.

Example two: Graph: $y=-x^{\mathrm{two}}-2x+\mathrm{iii}$ .

Solution:

Pace 1: Determine the y-intercept. To practice this, fix x = 0 and solve for y.

The y-intercept is (0, 3).

Step ii: Decide the x-intercepts. To do this, prepare y = 0 and solve for ten.

Here when y = 0, we obtain two solutions. At that place are two x-intercepts, (−three, 0) and (1, 0).

Stride 3: Determine the vertex. One way to do this is to utilize the equation for the line of symmetry, $x=-\frac{b}{2a}$ , to find the x-value of the vertex. In this example, a = −1 and b = −two:

Substitute −i into the original equation to find the corresponding y-value.

The vertex is (−ane, 4).

Step four: Determine extra points so that we take at to the lowest degree five points to plot. In this case, one other bespeak will suffice. Cull x = −2 and find the respective y-value.

Our fifth betoken is (−2, 3).

Step 5: Plot the points and sketch the graph. To epitomize, the points that nosotros have found are

y-intercept:	(0, 3)
x-intercept:	(−three, 0) and (i, 0)
Vertex:	(−1, iv)
Extra indicate:	(−2, 3)

Respond:

The parabola opens down. In general, use the leading coefficient to determine whether the parabola opens upward or downwardly. If the leading coefficient is negative, as in the previous example, and then the parabola opens downwards. If the leading coefficient is positive, then the parabola opens upward.

All quadratic equations of the grade $y=a{x}^{\mathrm{two}}+bx+c$ have parabolic graphs with y-intercept (0,c). Yet, not all parabolas accept ten intercepts.

Case 3: Graph: $y=\mathrm{ii}{x}^2+4\mathrm{10}+5$ .

Solution: Because the leading coefficient ii is positive, notation that the parabola opens up. Here c = v and the y-intercept is (0, v). To find the x-intercepts, set y = 0.

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 existent solutions. Because there are no existent solutions, there are no x-intercepts. Next, we determine the x-value of the vertex.

Given that the x-value of the vertex is −i, substitute into the original equation to find the corresponding y-value.

The vertex is (−1, 3). So far, we take merely two points. To determine 3 more than, cull some x-values on either side of the line of symmetry, ten = −ane. Here we cull x-values −three, −ii, and one.

To summarize, we have

y-intercept:	(0, five)
x-intercepts:	None
Vertex:	(−1, three)
Extra points:	(−3, xi), (−2, five), (1, 11)

Plot the points and sketch the graph.

Answer:

Example 4: Graph: $y=-2x^2+12x-18$ .

Solution: Notation that a = −2: the parabola opens downwards. Since c = −18, the y-intercept is (0, −18). To discover the 10-intercepts, set y = 0.

Solve by factoring.

Hither 10 = iii is a double root, and so there is only ane x-intercept, (iii, 0). From the original equation, a = −2, b = 12, and c = −18. The x-value of the vertex tin can be calculated every bit follows:

Given that the x-value of the vertex is three, substitute into the original equation to detect the corresponding y-value.

Therefore, the vertex is (iii, 0), which happens to exist the same point as the x-intercept. Then far, we have only two points. To determine three more, choose some x-values on either side of the line of symmetry, 10 = 3 in this case. Cull x-values 1, 5, and vi.

To summarize, we have

y-intercept:	(0, −18)
x-intercept:	(3, 0)
Vertex:	(3, 0)
Extra points:	(1, −8), (5, −8), (6, −18)

Plot the points and sketch the graph.

Answer:

Case 5: Graph: $y=x^{\mathrm{ii}}-2x-\mathrm{ane}$ .

Solution: Since a = ane, the parabola opens upward. Furthermore, c = −1, and so the y-intercept is (0, −ane). To observe the ten-intercepts, ready y = 0.

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 in that location are two 10-intercepts:

Estimate values using a calculator:

Use the approximate answers to place the ordered pair on the graph. However, we will nowadays the exact x-intercepts on the graph. Next, find the vertex.

Given that the 10-value of the vertex is 1, substitute into the original equation to find the corresponding y-value.

The vertex is (ane, −2). Nosotros need i more point.

To summarize, we accept

y-intercept:	(0, −one)
ten-intercepts:	( $1-\sqrt{\mathrm{two}}$ , 0) and ( $1+\sqrt{2}$ , 0)
Vertex:	(1, −2)
Actress point:	(2, −1)

Plot the points and sketch the graph.

Reply:

Effort this! Graph: $y=9x^2-5$ .

Respond:

Finding the Maximum and Minimum

It is often useful to observe the maximum and/or minimum values of functions that model real-life applications. To observe these important values given a quadratic part, we use the vertex. If the leading coefficient a is positive, then the parabola opens upward and there will be a minimum y-value. If the leading coefficient a is negative, then the parabola opens downward and in that location will be a maximum y-value.

Instance six: Determine the maximum or minimum: $y=-4x^{\mathrm{ii}}+24x-35$ .

Solution: Since a = −4, we know that the parabola opens downwardly and there will be a maximum y-value. To find information technology, nosotros kickoff detect the x-value of the vertex.

The 10-value of the vertex is 3. Substitute this value into the original equation to find the corresponding y-value.

The vertex is (iii, 1). Therefore, the maximum y-value is one, which occurs when x = 3, as illustrated below:

Notation

The graph is not required to answer this question.

Answer: The maximum is ane.

Instance 7: Determine the maximum or minimum: $y=4x^{\mathrm{two}}-32x+62$ .

Solution: Since a = +4, the parabola opens upward and at that place is a minimum y-value. Begin by finding the 10-value of the vertex.

Substitute ten = 4 into the original equation to find the respective y-value.

The vertex is (four, −2). Therefore, the minimum y-value of −two occurs when x = 4, every bit illustrated beneath:

Answer: The minimum is −2.

Try this! Determine the maximum or minimum: $y={\left(x-3\right)}^2-\mathrm{ix}$ .

Answer: The minimum is −ix.

A parabola, opening up or downwards (equally opposed to sideways), defines a function and extends indefinitely to the correct and left as indicated past the arrows. Therefore, the domain (the fix of x-values) consists of all existent numbers. However, the range (the prepare of y-values) is divisional by the y-value of the vertex.

Example 8: Decide the domain and range: $y={\mathrm{10}}^2-4x+3$ .

Solution: Get-go, note that since $a=1$ is positive, the parabola opens upward. Hence there will be a minimum y-value. To discover that value, observe the x-value of the vertex:

So substitute into the equation to find the corresponding y-value.

The vertex is (2, −1). The range consists of the set of y-values greater than or equal to the minimum y-value −1.

Answer: Domain: R = (−∞, ∞); range: [−1, ∞)

Example 9: The superlative in anxiety of a projectile is given past the office $h(t)=-16t^{\mathrm{two}}+72t$ , where t represents the time in seconds after launch. What is the maximum height reached by the projectile?

Solution: Hither $a=-16$ , and the parabola opens downwards. Therefore, the y-value of the vertex determines the maximum height. Brainstorm by finding the ten-value of the vertex:

The maximum height will occur in nine/4 = 2¼ seconds. Substitute this time into the role to determine the acme attained.

Answer: The maximum height of the projectile is 81 feet.

Finding the Vertex by Completing the Foursquare

In this department, we demonstrate an alternating arroyo for finding the vertex. Any quadratic equation $y=a{\mathrm{ten}}^2+b\mathrm{ten}+c$ can be rewritten in the form

In this form, the vertex is (h,k).

Instance 10: Determine the vertex: $y=-\mathrm{iv}{\left(x-\mathrm{iii}\right)}^2+1$ .

Solution: When the equation is in this form, we tin read the vertex straight from the equation.

Here h = 3 and k = 1.

Answer: The vertex is (3, 1).

Example 11: Determine the vertex: $y=\mathrm{ii}{\left(\mathrm{10}+3\right)}^{\mathrm{two}}-2$ .

Solution: Rewrite the equation every bit follows before determining h and k.

Here h = −three and one thousand = −2.

Reply: The vertex is (−3, −2).

Oft the equation is not given in this form. To obtain this form, complete the foursquare.

Example 12: Rewrite in $y=a{\left(x-h\right)}^2+k$ form and determine the vertex: $y=x^2+\mathrm{four}x+9$ .

Solution: Brainstorm by making room for the constant term that completes the square.

The idea is to add and subtract the value that completes the square, ${\left(\frac{b}{\mathrm{ii}}\right)}^2$ , and and so factor. In this case, add together and subtract ${\left(\frac{4}{2}\right)}^2={\left(2\right)}^2=4$ .

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 course, we can hands make up one's mind the vertex.

Here h = −two and g = five.

Answer: The vertex is (−2, 5).

If in that location is a leading coefficient other than one, and then nosotros must showtime gene out the leading coefficient from the get-go two terms of the trinomial.

Example 13: Rewrite in $y=a{\left(x-h\right)}^2+k$ course and make up one's mind the vertex: $y=2{\mathrm{ten}}^2-4\mathrm{10}+8$ .

Solution: Since a = ii, factor this out of the kickoff two terms in guild to consummate the square. Get out room inside the parentheses to add a constant term.

At present employ −2 to determine the value that completes the foursquare. In this case, ${\left(\frac{-2}{2}\right)}^2={\left(-\mathrm{ane}\right)}^2=1$ . Add and subtract 1 and cistron as follows:

In this course, nosotros tin can easily decide the vertex.

Here h = 1 and k = half dozen.

Answer: The vertex is (i, 6).

Try this! Rewrite in $y=a{\left(\mathrm{10}-h\right)}^{\mathrm{two}}+k$ form and determine the vertex: $y=-\mathrm{two}{x}^2-12\mathrm{ten}+3$ .

Answer: $y=-\mathrm{ii}{\left(x+3\right)}^2+21$ ; vertex: (−three, 21)

Key Takeaways

• The graph of whatsoever quadratic equation $y=a{x}^{\mathrm{two}}+b\mathrm{ten}+c$ , where a, b, and c are real numbers and $a\ne 0$ , is called a parabola.
• When graphing parabolas, detect the vertex and y-intercept. If the x-intercepts exist, notice those as well. Also, be sure to find ordered pair solutions on either side of the line of symmetry, $\mathrm{10}=-\frac{b}{2a}$ .
• Utilize 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 downwards.
• The vertex of any parabola has an x-value equal to $-\frac{b}{2a}$ . After finding the x-value of the vertex, substitute information technology into the original equation to find the corresponding y-value. This y-value is a maximum if the parabola opens down, and information technology is a minimum if the parabola opens up.
• The domain of a parabola opening upwards or downward consists of all real numbers. The range is bounded by the y-value of the vertex.
• An alternate approach to finding the vertex is to rewrite the quadratic equation in the form $y=a{\left(x-h\right)}^2+k$ . When in this form, the vertex is (h,k) and can exist read directly from the equation. To obtain this form, accept $y=a{x}^2+bx+c$ and complete the square.

Topic Exercises

Office A: The Graph of Quadratic Equations

Does the parabola open upward or down? Explain.

i. $y={\mathrm{10}}^2-9x+\mathrm{xx}$

2. $y=x^2-12\mathrm{ten}+32$

3. $y=-\mathrm{two}{\mathrm{ten}}^2+\mathrm{five}x+12$

4. $y=-6x^2+\mathrm{xiii}x-\mathrm{half-dozen}$

five. $y=64-{\mathrm{10}}^2$

6. $y=-3x+9{\mathrm{ten}}^2$

Determine the 10 - and y -intercepts.

7. $y=x^2+4x-12$

viii. $y=x^{\mathrm{two}}-13\mathrm{10}+12$

9. $y=2x^2+\mathrm{five}\mathrm{ten}-\mathrm{iii}$

x. $y=3{\mathrm{ten}}^{\mathrm{two}}-4\mathrm{10}-4$

xi. $y=-\mathrm{five}{\mathrm{10}}^2-\mathrm{three}x+\mathrm{two}$

12. $y=-6{\mathrm{ten}}^{\mathrm{ii}}+\mathrm{eleven}x-\mathrm{four}$

13. $y=\mathrm{iv}{x}^2-25$

14. $y=9x^{\mathrm{ii}}-49$

xv. $y=x^{\mathrm{ii}}-x+1$

16. $y=5x^2+15x$

Observe the vertex and the line of symmetry.

17. $y=-x^2+10\mathrm{10}-34$

18. $y=-{\mathrm{10}}^{\mathrm{ii}}-6\mathrm{10}+1$

nineteen. $y=-4x^{\mathrm{ii}}+12\mathrm{10}-7$

xx. $y=-9x^{\mathrm{two}}+6x+2$

21. $y=4x^{\mathrm{ii}}-1$

22. $y=x^{\mathrm{ii}}-\mathrm{sixteen}$

Graph. Find the vertex and the y -intercept. In addition, find the x -intercepts if they exist.

23. $y=x^2-\mathrm{ii}x-8$

24. $y=x^2-4x-5$

25. $y=-{\mathrm{ten}}^2+4\mathrm{10}+12$

26. $y=-{\mathrm{ten}}^2-2x+15$

27. $y=x^2-10x$

28. $y=x^2+\mathrm{viii}x$

29. $y=x^{\mathrm{ii}}-9$

30. $y=x^{\mathrm{two}}-25$

31. $y=\mathrm{i}-{\mathrm{10}}^{\mathrm{two}}$

32. $y=4-{\mathrm{10}}^2$

33. $y=x^{\mathrm{ii}}-2x+1$

34. $y={\mathrm{10}}^{\mathrm{two}}+4x+\mathrm{four}$

35. $y=-\mathrm{iv}{\mathrm{ten}}^2+12x-9$

36. $y=-4{\mathrm{ten}}^2-\mathrm{four}x+\mathrm{iii}$

37. $y={\mathrm{10}}^2-2$

38. $y={\mathrm{ten}}^{\mathrm{ii}}-3$

39. $y=-\mathrm{iv}{x}^2+\mathrm{iv}x-3$

xl. $y=4{\mathrm{ten}}^2+4x+3$

41. $y=x^2-2\mathrm{10}-\mathrm{two}$

42. $y={\mathrm{10}}^2-\mathrm{six}\mathrm{10}+6$

43. $y=-2x^2+\mathrm{half-dozen}\mathrm{10}-3$

44. $y=-4x^2+4x+1$

45. $y=x^2+\mathrm{three}x+4$

46. $y=-{\mathrm{ten}}^{\mathrm{ii}}+3\mathrm{10}-4$

47. $y=-2x^{\mathrm{ii}}+3$

48. $y=-\mathrm{two}{x}^2-1$

49. $y=\mathrm{two}{x}^{\mathrm{two}}+4x-3$

l. $y=3x^2+2\mathrm{10}-2$

Function B: Maximum or Minimum

Determine the maximum or minimum y -value.

51. $y=-x^{\mathrm{ii}}-6\mathrm{10}+1$

52. $y=-x^{\mathrm{ii}}-4\mathrm{ten}+8$

53. $y=25x^2-10x+\mathrm{five}$

54. $y=16{\mathrm{ten}}^2-24\mathrm{10}+\mathrm{vii}$

55. $y=-{\mathrm{ten}}^2$

56. $y=1-9x^{\mathrm{two}}$

57. $y=20x-10x^2$

58. $y=12x+4x^2$

59. $y=3x^{\mathrm{two}}-\mathrm{four}x-2$

threescore. $y=\mathrm{half-dozen}{\mathrm{10}}^{\mathrm{two}}-8\mathrm{ten}+5$

Given the following quadratic functions, determine the domain and range.

61. $f(x)=\mathrm{three}{\mathrm{10}}^2+\mathrm{xxx}\mathrm{ten}+50$

62. $f(x)=\mathrm{five}{x}^{\mathrm{two}}-10x+1$

63. $g(\mathrm{ten})=-\mathrm{two}{x}^{\mathrm{ii}}+4x+\mathrm{i}$

64. $g(\mathrm{ten})=-7{\mathrm{ten}}^2-\mathrm{fourteen}\mathrm{ten}-9$

65. The acme in feet reached by a baseball game tossed upwardly at a speed of 48 feet/2d from the basis is given by the function $h\left(t\right)=-\mathrm{sixteen}{t}^{\mathrm{ii}}+48t$ , where t represents time in seconds. What is the baseball game's maximum tiptop and how long will it take to reach that superlative?

66. The tiptop of a projectile launched directly upward from a mound is given by the role $h(t)=-\mathrm{xvi}{t}^{\mathrm{ii}}+96t+4$ , where t represents seconds subsequently launch. What is the maximum height?

67. The profit in dollars generated by producing and selling x custom lamps is given by the function $P(\mathrm{ten})=-\mathrm{x}{\mathrm{10}}^2+800x-\mathrm{12,000}$ . What is the maximum turn a profit?

68. The revenue in dollars generated from selling a particular detail is modeled past the formula $R(\mathrm{ten})=100\mathrm{ten}-0.0025x^2$ , where x represents the number of units sold. What number of units must be sold to maximize revenue?

69. The boilerplate number of hits to a radio station website is modeled by the formula $f(\mathrm{ten})=450t^2-\mathrm{iii,600}t+\mathrm{8,000}$ , where t represents the number of hours since 8:00 a.yard. 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 $\mathrm{Five}(t)=125t^2-\mathrm{3,000}t+\mathrm{22,000}$ , where t represents the number of years since it was purchased. Determine the minimum value of the car.

71. The daily product costs in dollars of a textile manufacturing company producing custom uniforms is modeled by the formula $C(x)=0.02x^2-20\mathrm{ten}+\mathrm{10,000}$ , where ten represents the number of uniforms produced.

a. How many uniforms should exist 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 past the formula $A=14w-{\mathrm{westward}}^2$ , where w represents the width in anxiety. Decide the width that produces the maximum area.

Part C: Vertex by Completing the Square

Determine the vertex.

73. $y=-{\left(x-5\right)}^2+3$

74. $y=-2{\left(x-\mathrm{one}\right)}^{\mathrm{two}}+7$

75. $y=\mathrm{five}{\left(x+\mathrm{i}\right)}^2+6$

76. $y=\mathrm{three}{\left(x+4\right)}^2+10$

77. $y=-5{\left(x+8\right)}^2-1$

78. $y={\left(x+2\right)}^2-\mathrm{five}$

Rewrite in $y=a{\left(x-h\right)}^{\mathrm{ii}}+k$ form and determine the vertex.

79. $y=x^{\mathrm{ii}}-\mathrm{xiv}\mathrm{10}+24$

80. $y=x^2-12\mathrm{10}+40$

81. $y=x^2+\mathrm{four}\mathrm{ten}-12$

82. $y=x^{\mathrm{ii}}+\mathrm{six}x-\mathrm{ane}$

83. $y=\mathrm{ii}{\mathrm{ten}}^2-12x-3$

84. $y=3x^2-6x+5$

85. $y=-x^{\mathrm{two}}+\mathrm{xvi}x+17$

86. $y=-{\mathrm{10}}^2+\mathrm{ten}x$

Graph.

87. $y=x^{\mathrm{two}}-1$

88. $y=x^2+1$

89. $y={\left(x-1\right)}^2$

90. $y={\left(x+\mathrm{i}\right)}^2$

91. $y={\left(\mathrm{10}-4\right)}^2-\mathrm{nine}$

92. $y={\left(\mathrm{ten}-1\right)}^2-4$

93. $y=-2{\left(x+1\right)}^{\mathrm{ii}}+8$

94. $y=-3{\left(x+2\right)}^2+12$

95. $y=-5{\left(x-1\right)}^2$

96. $y=-{\left(x+2\right)}^2$

97. $y=-4{\left(x-1\right)}^2-2$

98. $y=9{\left(\mathrm{ten}+1\right)}^2+2$

99. $y={\left(x+5\right)}^2-\mathrm{fifteen}$

100. $y=2{\left(x-5\right)}^2-3$

101. $y=-2{\left(x-4\right)}^2+22$

102. $y=2{\left(x+3\right)}^2-13$

Office D: Give-and-take Lath

103. Write downward your programme for graphing a parabola on an exam. What will you be looking for and how will y'all present your answer? Share your plan on the discussion board.

104. Why is any parabola that opens upwardly or downward a role? Explain to a classmate how to determine the domain and range.

Answers

one: Upwardly

3: Downward

5: Downward

7: x-intercepts: (−half-dozen, 0), (2, 0); y-intercept: (0, −12)

9: x-intercepts: (−3, 0), (1/2, 0); y-intercept: (0, −three)

11: x-intercepts: (−1, 0), (2/five, 0); y-intercept: (0, ii)

xiii: x-intercepts: (−5/2, 0), (5/2, 0); y-intercept: (0, −25)

15: x-intercepts: none; y-intercept: (0, 1)

17: Vertex: (5, −9); line of symmetry: $x=5$

nineteen: Vertex: (three/2, ii); line of symmetry: $x=3/2$

21: Vertex: (0, −ane); 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 = −x/iii

61: Domain: R; range: $\left[-25,\infty \right)$

63: Domain: R; range: $\left(-\infty ,3\right]$

65: The maximum elevation of 36 feet occurs after 1.5 seconds.

67: $four,000

69: 12:00 p.chiliad.

71: a. 500 uniforms; b. $5,000

73: (5, 3)

75: (−one, 6)

77: (−8, −one)

79: $y={\left(\mathrm{10}-7\right)}^{\mathrm{two}}-25$ ; vertex: (7, −25)

81: $y={\left(x+2\right)}^{\mathrm{two}}-16$ ; vertex: (−2, −sixteen)

83: $y=\mathrm{ii}{\left(x-3\right)}^2-21$ ; vertex: (3, −21)

85: $y=-{\left(\mathrm{ten}-8\right)}^2+81$ ; vertex: (8, 81)

87:

89:

91:

93:

95:

97:

99:

101:

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Source: https://saylordotorg.github.io/text_elementary-algebra/s12-05-graphing-parabolas.html

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