1. Liv1234

2. Michele_Laino

the new function g(x) has to be equal to the old function f at point x+1

3. Liv1234

What does that mean?

4. Michele_Laino

in formula, my statements means: g(x)=f(x+1)

5. Liv1234

So kind of like for example y= mx+b?

6. Michele_Laino

in that case, we can write this: g(x9=m(x+1)+b

7. Michele_Laino

oops.. g(x)=m(x+1)+b

8. Liv1234

By the way, sorry if I seem slow with this, I'm just really bad with math.

9. Michele_Laino

no worries! :)

10. Liv1234

So, g(x)=m(x+1)+b

11. Michele_Laino

yes!

12. Liv1234

How would I solve the answer with that?

13. Michele_Laino

I think you have to do the multiplication, like below: g(x)=mx+m+b

14. Liv1234

How do I do the multiplication with this equation?

15. Michele_Laino

we can do the multiplication above using the distributive property of multiplication over addition

16. Liv1234

I think I remember how to do that, is there a formula?

17. Liv1234

So, (sorry to ask again) what numbers would be put in?

18. Michele_Laino

the general formula, is: $\left( {a + b} \right)c = ac + bc,\quad \forall a,b,c \in \mathbb{R}$

19. Michele_Laino

you have to know the values of the coefficients m and b

20. Liv1234

I feel really stupid to keep asking for help, but how would I find the coefficients?

21. Michele_Laino

you can retrieve the value of those coefficients, from the text of your exercise

22. Liv1234

The only number I had from the exercise is "1"

23. Michele_Laino

in that case your exercise is not a numeric exercise

24. Liv1234

Hang on, let me repost the question

25. Michele_Laino

ok!

26. Liv1234

27. Liv1234

Okay, that's the question

28. Michele_Laino

your answer, is: $g\left( x \right) = m\left( {x + 1} \right) + b = mx + m + b$ and we have completed your exercise

29. Liv1234

Wait, so what option would I choose? I'm confused.

30. Michele_Laino

the first option

31. Liv1234

Can you help me with another question?

32. Michele_Laino

ok!

33. Liv1234

34. Michele_Laino

when we make a reflection with respect to the x-axis, then we can write the new function g(x) as below: $g\left( x \right) = - f\left( x \right)$

35. Liv1234

So, how would I solve?

36. Michele_Laino

it is simple, you have to replace f(x) with its formula: $g\left( x \right) = - f\left( x \right) = - \left( {{x^2} + 5} \right) = ...?$

37. Liv1234

Would the answer be g(x)= -x^2+5?

38. Michele_Laino

not exactly, since when you cancel the parentheses, you get this: $g\left( x \right) = - f\left( x \right) = - \left( {{x^2} + 5} \right) = - {x^2} - 5$

39. Liv1234

I'm confused D:

40. Michele_Laino

every term inside the parentheses change its sign

41. Michele_Laino

so we can write this: $- \left( {{x^2} + 5} \right) = - {x^2} - 5$

42. Liv1234

So, the first option?

43. Michele_Laino

yes!

44. Liv1234

Can you help me with another one?

45. Michele_Laino

ok!

46. Liv1234

47. Michele_Laino

here we have to consider all points whose y-coordinate is less the y coordinate of the points which belong to the straight line

48. Liv1234

Let me guess, it's the first option right?

49. Liv1234

Or is it another one?

50. Michele_Laino

please wait, first we have to write the equation of the straight line

51. Liv1234

Okay, so how would we do that?

52. Liv1234

I know that the line is at -6 and 5

53. Michele_Laino

your line passes at points (-5,0) and (0,6), so its equation is: $\frac{{y - 0}}{{6 - 0}} = \frac{{x + 5}}{{0 + 5}}$

54. Liv1234

It actually passes at -6 and 5 though.

55. Michele_Laino

which can be simplified to this form: $6x - 5y = - 30$

56. Liv1234

So, what option would I choose?

57. Michele_Laino

please wait, I think that your line passes at points (-6,0) and (0,5), so its equation is: $5x - 6y = - 30$

58. Liv1234

So, would it be C or D?

59. Michele_Laino

so the shaded part of cartesian plane is represented by this inequality: $5x - 6y \geqslant - 30$

60. Liv1234

So, D?

61. Liv1234

Can you help me with another one please?

62. Michele_Laino

more explanation: the y-coordinate of the points of the shaded region has to be less than the y-coordinate of the point of the straight line, so we can write this: $y \leqslant \frac{{5x - 30}}{6}$

63. Liv1234

Can you help me with this question?

64. Liv1234

65. Michele_Laino

we can write this: $g\left( x \right) = 5\left( { - 7{x^2}} \right) + 5 = 5f\left( x \right) + 5$

66. Liv1234

I think the answer is D

67. Michele_Laino

yes! correct!

68. Liv1234

Yay! Can you help me with another one please?

69. Michele_Laino

ok!

70. Liv1234

71. Liv1234

Okay, that's the question.

72. Michele_Laino

the y-coordinates of the new function g(x) have to be six time greater than the respectively y-coordinate of the old function f(x)

73. Liv1234

So, would the answer be C?

74. Michele_Laino

no, I meant six as a factor

75. Liv1234

Oh, so would it be A? Because it has multiplication?

76. Michele_Laino

correct!

77. Liv1234

Can you help me with another one?

78. Michele_Laino

please wait, since another student has requested my help

79. Liv1234

It's a quick question though

80. Michele_Laino

ok!

81. Liv1234

82. Michele_Laino

-7 means shifting down by 7 units furthermore, as we can see in your first exercise, x+3 means shifting to left

83. Liv1234

So, would it be C?

84. Liv1234

Oh, wait, B?

85. Michele_Laino

no, since the vertical shift is by 7 units down and the horizontal shift is by 3 units left

86. Liv1234

So, B?

87. Michele_Laino

correct! it is option B

88. Liv1234

Do you think you could help me with another?

89. Michele_Laino