## anonymous one year ago f(x) = -16t^2 + 96t what is the x and y intercept

1. freckles

where is the function of x? is t suppose to be x?

2. freckles

well I mean that could be a function of x it would just be a constant function of x

3. anonymous
4. Vocaloid

|dw:1436976311744:dw|

5. Vocaloid

|dw:1436976350942:dw|

6. Vocaloid

the x-intercepts are the points where the function is equal to 0 the y-intercept is the point where x = 0

7. anonymous

is the x intercept 1,6?

8. Vocaloid

yes, or more specifically (1,0) and (6,0)

9. anonymous

also could you help me with a few more questions

10. anonymous

Find the vertex algebraically. Write as an ordered pair.

11. anonymous

What is the maximum height of the ball? Write your answer as an integer.

12. freckles

(1,0) isn't on the graph though :p

13. anonymous

so it would 6,0

14. freckles

(0,0) and (6,0) if we are talking about the function in the graph

15. anonymous

okay

16. freckles

$f(x) = -16t^2 + 96t$ so I guess that that one x was suppose to be a t :p

17. anonymous

oh

18. anonymous

Find the vertex algebraically. Write as an ordered pair. would the vertex 6,0 ?

19. freckles

wait and it looks like they called the function h(t) instead of f(t)

20. freckles

but whatever

21. freckles

the vertex would be that peak point

22. freckles

if you wanted to find it using that graph you could observe that the vertex occurs midway between the x-intercepts

23. anonymous

which is 3?

24. freckles

I will also show you how to find it algebraically since you asked

25. freckles

but first using the graph the vertex appears to be at (3,a little above 140) I will let you approximate that y value

26. freckles

but anyways algebraically ...

27. freckles

$y=at^2+bt+c \\ \text{ we want to put this in vertex form } \\$ $y-c =at^2+bt \text{ I subtract } c \text{ on both sides } \\ \frac{y-c}{a} =t^2+\frac{b}{a} t \text{ I divided both sides by } a \\ \frac{y-c}{a} +(\frac{b}{2a})^2=t^2+\frac{b}{a}t +(\frac{b}{2a})^2 \\ \text{ I threw in a number on both sides so I could } \\ \text{ complete the square on one side } \\ \frac{y-c}{a}+(\frac{b}{2a})^2= (t+\frac{b}{2a})^2 \\ \text{ now \let's re-isolate } y \\ \text{ subtract} (\frac{b}{2a})^2 \text{ on both sides } \\ \frac{y-c}{a}=(t+\frac{b}{2a})^2-(\frac{b}{2a})^2 \\ \text{ multiply } a \text{ on both sides } \\ y-c=a(t+\frac{b}{2a})^2-a (\frac{b}{2a})^2 \\ \text{ add } c \text{ on both sides } \\ y=a(t+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c$ the vertex is given by: $(-\frac{b}{2a},-a(\frac{b}{2a})^2+c)$

28. anonymous

3o would the vertex be 3,14 or 3,140?

29. freckles

neither

30. freckles

look at the graph you should see where that mountain's peak is just a hair above 140 the y value isn't exactly 140 it is just above 140

31. anonymous

160?

32. freckles

160 is too high

33. freckles

it is closer to 140 than 160

34. freckles

|dw:1436977378401:dw|

35. freckles

|dw:1436977410220:dw|

36. anonymous

120

37. freckles

how is 120 just a little above 140 ?

38. freckles

you can find the y-coordinate of the vertex exactly by using the steps I gave you above

39. anonymous

sorry I was confused

40. freckles

but you can approximate it using the visual it looks like it could be maybe less than 150 but greater than 140 and I say less than 150 because the peak doesn't reach halfway between 140 and 160 it is less than halfway I would approximate saying 142 but that could be off a little because I did find this visually and not algebraically as the instructions|dw:1436977561263:dw| suggested

41. freckles

|dw:1436977596987:dw| so you know the vertex is gonna be somewhere close to (3,142)

42. freckles

you can find the answer algebraically using the method above I suggested

43. freckles

this one: $y-c =at^2+bt \text{ I subtract } c \text{ on both sides } \\ \frac{y-c}{a} =t^2+\frac{b}{a} t \text{ I divided both sides by } a \\ \frac{y-c}{a} +(\frac{b}{2a})^2=t^2+\frac{b}{a}t +(\frac{b}{2a})^2 \\ \text{ I threw in a number on both sides so I could } \\ \text{ complete the square on one side } \\ \frac{y-c}{a}+(\frac{b}{2a})^2= (t+\frac{b}{2a})^2 \\ \text{ now \let's re-isolate } y \\ \text{ subtract} (\frac{b}{2a})^2 \text{ on both sides } \\ \frac{y-c}{a}=(t+\frac{b}{2a})^2-(\frac{b}{2a})^2 \\ \text{ multiply } a \text{ on both sides } \\ y-c=a(t+\frac{b}{2a})^2-a (\frac{b}{2a})^2 \\ \text{ add } c \text{ on both sides } \\ y=a(t+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c$

44. anonymous

I got 141 ?

45. anonymous

though I think it may be -3 I'm not sure

46. freckles

how did you get 141?

47. freckles

48. anonymous

I'm still using the graph for it

49. freckles

oh okay well don't forget it does say to find the vertex algebraically

50. freckles

using a graph to obtain the vertex is not finding it algebraically

51. anonymous

oh

52. anonymous

so it's not -3

53. freckles

are you talking about the x-coordinate of the vertex?

54. anonymous

vertex

55. freckles

no -3 is not the x-coordinate

56. freckles

|dw:1436978591061:dw|

57. anonymous

141 ?

58. freckles

the vertex is pretty close to (3,142) or I guess you could say (3,141) but we don't know the vertex exactly until you find it algebraically

59. freckles

it could be (3,144)

60. freckles

or (3,143)

61. freckles

but to find out we need to put your quadratic in vertex form

62. freckles

I gave a general way to do that above

63. freckles

|dw:1436979098053:dw| please look at this graph... |dw:1436979115849:dw|

64. freckles

again this is not finding the vertex algebraically this is approximating the vertex with the help of visual aid you can find the exact vertex by putting the quadratic into vertex form you will need to complete the square

65. freckles

$y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+?)+16(?)$ what do you need to add inside the ( ) to complete the square

66. anonymous

1 ?

67. freckles

no recall that: $x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2$

68. freckles

the number in front of t is -6 let's call k, -6 so if k=-6 then k/2=?

69. anonymous

it would -6/2

70. freckles

$y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2$ right can you finish putting into vertex form

71. freckles

by the way I didn't put the - in because (6/2)^2 is the same as (-6/2)^2 but you could have

72. anonymous

73. freckles

no

74. freckles

we already know we are suppose to get something close to (3,142) because this what we found visually

75. freckles

so I don't know what -6 is suppose to be

76. anonymous

would it 3,2

77. freckles

but anyways I was asking if you could finish putting into the vertex form

78. freckles

i have to ask are you playing with me?

79. freckles

like you keep giving me these y values that aren't closer to 140

80. anonymous

no I'm not i'm really stuck on this equation

81. freckles

I have shown you visually the y-value is a little above 140 we know the x-value to be exactly 3 the vertex is definitely somewhere close to (3, 142)

82. anonymous

83. freckles

$y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2 \\ y=-16(t-3)^2+16(\frac{6}{2})^2$ there I completed the square for you

84. freckles

as we see the x-coordinate of the vertex is 3 just evaluate 16(6/2)^2 to find the y coordinate of the vertex

85. freckles

@Aliypop yes I see you said 141 as approximate but then you asked me if it was 2,14,... and so on...

86. anonymous

I confused on what the x value is

87. freckles

the x-coordinate of the vertex is 3 we have found this both visually and algebraically now

88. freckles

the y-coordinate of the vertex is given by 16(6/2)^2 which we have approximated to be somewhere just above 140

89. anonymous

so is it (3,3)

90. freckles

no...

91. freckles

again 3 is nowhere near 140

92. freckles

the x-coordinate of the vertex is 3 the y-coordinate of the vertex is something just above 140 we have found this information visually we have also found algebraically the x-coordinate of the vertex is 3 the y-coordinate of the vertex is 16(6/2)^2

93. freckles

$y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2 \\ y=-16(t-3)^2+16(\frac{6}{2})^2 \\ \text{ this says the vertex is } (3,16(\frac{6}{2})^2)$

94. freckles

all you have to do is finishing simplifying

95. freckles

do you need help doing 16(6/2)^2?

96. freckles

6 divided by 2 is 3 3 squared is 9 16*9 is...

97. anonymous

144

98. freckles

yes $y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2 \\ y=-16(t-3)^2+16(\frac{6}{2})^2 \\ \text{ this says the vertex is } (3,16(\frac{6}{2})^2) =(3,144)$

99. freckles

the vertex is exactly (3,144)

100. freckles

as we see this makes sense from our visual answer because we did say the vertex had x-coordinate 3 and y-coordinate of something just above 140

101. anonymous

okay

102. anonymous

What is the maximum height of the ball? Write your answer as an integer. this was the other question i need help on

103. freckles

you actually already found this when you found the vertex the vertex is the highest point and you know the y-value represents the height

104. anonymous

so it would 144

105. freckles

|dw:1436980560407:dw| yes the height is 144 and the time in which the height is 144 is t=3 sec if the time unit is second

106. freckles

though usually you would say 144 units where units is some kind of distance measurement but that isn't really given here