## anonymous one year ago grr

1. Michele_Laino

we know that your function has to be a polynomial of degree 2 from your drawing we see that x=6 and x=-6 are two zeroes of the requested polynomial

2. anonymous

Oh! So would it be -1(x+6)(x-6) ??

3. Michele_Laino

the simpler polynomial which satisfies those requisites is: $y = \left( {x - 6} \right)\left( {x + 6} \right)$ even if, we have to check if my function contains the point (0,36) does my function pass at point (0,36)?

4. Michele_Laino

yes! that's right!

5. anonymous

Ok so then I got -x^2+36?

6. Michele_Laino

correct!

7. anonymous

There is a 2nd part to this question, I'm not getting

8. anonymous

I know, I'm lost too:)

9. Michele_Laino

you have to create a table for a straight line like this one: |dw:1438706644480:dw|

10. Michele_Laino

11. Michele_Laino

oops.. choose*

12. anonymous

um 3?

13. Michele_Laino

you have to choose 2 values for x-coordinate

14. Michele_Laino

between -6 and 6

15. anonymous

oh okay then 5

16. Michele_Laino

x1=5 and x2=?

17. anonymous

11?

18. Michele_Laino

no, your value has to be less than 6 and grater than -6

19. Michele_Laino

x=5 is right! the other value can be x=2

20. Michele_Laino

what do you think?

21. anonymous

yes, i agree

22. Michele_Laino

ok!

23. Michele_Laino

now we have to compute the corresponding y-value, so, if x=5 then: y=-5^2+36=-25+36=11

24. Michele_Laino

whereas if x=1, then: y=-1^2+25=...?

25. anonymous

I see, I see:)

26. anonymous

24

27. Michele_Laino

sorry if x=1, then: y=-1^2+36=35 we have these points: (1,35) and (5,11)

28. anonymous

I agree:)

29. anonymous

so then we would have to make a chart, and another function right?

30. Michele_Laino

now: step#2 we have to write the equation of the straight line which passes at points (1,35) and (5,11)

31. Michele_Laino

for example we can use this equation: $\Large y - 35 = m\left( {x - 1} \right)$ where: $\Large m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{11 - 35}}{{5 - 1}} = ...?$

32. anonymous

-24/4

33. anonymous

-6

34. Michele_Laino

yes!

35. Michele_Laino

so, what is the equation of your straight line?

36. anonymous

um y=6x+1?

37. anonymous

i mean y=-6x

38. Michele_Laino

we have to substitute our value of m, into my equation above: $\Large y - 35 = - 6 \cdot \left( {x - 1} \right)$

39. Michele_Laino

40. anonymous

y-35=-6x+6

41. Michele_Laino

so: y=...?

42. anonymous

y=-6x-29

43. Michele_Laino

I got: Y=-6x+41 am I right?

44. anonymous

yes. whoops:)

45. Michele_Laino

ok! Now the exercise asks you at least four points

46. Michele_Laino

you have already two points: (1,35) and (5,11)

47. Michele_Laino

so you have to choose another 2 values for x, and compute the corresponding value for y. Please keep in mind that x has to be greater than -6 and less than 6

48. anonymous

49. Michele_Laino

whit those four points you have to create a table like this: |dw:1438708104245:dw|

50. Michele_Laino

ok! x=2 is good, so: y= -2^2+36=-4+36=...?

51. anonymous

32

52. Michele_Laino

so the third point is: (2,32)

53. Michele_Laino

now, we can choose x=-3

54. Michele_Laino

what is y?

55. anonymous

45

56. Michele_Laino

I got this: y=-(-3^2)+36=-9+36=27

57. anonymous

yes<3

58. anonymous

Yayyy! We did it! So just to make sure, the 2nd equation is y=-6x?

59. Michele_Laino

so, the requested table is: |dw:1438708531419:dw|

60. Michele_Laino

the second equation is: $\Large y = - 6x + 41$

61. anonymous

oh, so is it possible that we could have plug the points into that equation to get the points too?

62. Michele_Laino

63. Michele_Laino

I have made an error, at x=2 we have to compute the value of the y-coordinate of the line, namely: $\Large y = - 6 \cdot 2 + 41 = ...$

64. Michele_Laino

similarly at x=-3: $\Large y = - 6 \cdot 3 + 41 = ...$

65. anonymous

29

66. anonymous

so can we draw the chart again and see?

67. Michele_Laino

ok! so we got these points: (2,29) and (-3,23)

68. anonymous

|dw:1438708966376:dw|

69. anonymous

what would the domain/range be?

70. Michele_Laino

that's right!

71. Michele_Laino

the domain of the direction of the drone, it is all the real line

72. anonymous

so would the domain be -6 and 6 then?

73. Michele_Laino

it is the domain of the rainbow, I think

74. anonymous

ohhh

75. anonymous

Thanks for all your help!!!!! :D

76. Michele_Laino

please wait, I have made another error, the fourth point is: (-3,59)

77. anonymous

Good thing you caught that!

78. Michele_Laino

since we have: $\Large y = \left( { - 6} \right) \cdot \left( { - 3} \right) + 41 = 18 + 41 = 59$

79. Michele_Laino

the domain of the rainbow is the subsequent interval: (-6,6) whereas its range is (0,36)

80. anonymous

what would that represent?

81. Michele_Laino

the range is the set af possible values of the function -x^2+36 when x is inside the domain

82. anonymous

Is the linear function you created positive or negative? Explain.

83. Michele_Laino

I think that it represents the possible Rainbow heights with respect to the Earth's surface

84. Michele_Laino

the linear function which we have created has a negative slope, so I think it is negative

85. anonymous

What are the solutions or solution to the system of equations created? Explain what it or they represent.

86. anonymous

thanks so much:)

87. Michele_Laino

the requested solution is given by the coordinates of the intersection points, namely: (1,35) and (5,11)

88. Michele_Laino

:)