## monkey* one year ago Given f(x)=x^2+7 and g(x)=x-4/x. Find (g o f)(-1). Hint: (g o f)(-1)=g(f(-1))

1. freckles

so have you tried finding f(-1) and then finding g(put result of f(-1) here)

2. freckles

f(-1) tells us to use f(x)=x^2+7 we know to use f(x)=x^2+7 because of the f in f(-1)

3. freckles

f(x)=x^2+7 we can see here the input is the thing in f( ) example: f(a)=a^2+7 f(a+1)=(a+1)^2+7 f(input)=(input)^2+7 another example; f(2)=2^2+7 can you find f(-1)=?

4. monkey*

I'm a little lost, sorry.

5. freckles

ok do you see that f(x)=x^2+7 and you are also to find f(-1)?

6. freckles

f(x)=x^2+7 since we want to find f(-1) we need to replace those x's with -1

7. monkey*

Yes.

8. freckles

so can you find f(-1)?

9. freckles

|dw:1440459828779:dw|

10. freckles

let me know when you have found f(-1)

11. monkey*

Would it be f(-1)-1^2+7?

12. freckles

well there should be an equal sign somewhere there

13. freckles

|dw:1440460061483:dw|

14. freckles

can you simplify (-1)^2+7?

15. monkey*

8?

16. freckles

yes f(-1)=(-1)^2+7=(-1)(-1)+7=1+7=8 so f(-1)=8 so f(-1) is 8 now we had to find g(f(-1)) we just found that inside thing the f(-1) thing to be 8 replace f(-1) with 8 you are now really wanting to find g(8)

17. freckles

g(8) tells us to use the expression named g

18. freckles

g looks likes g(x)=x-4/x

19. freckles

just replace the old input x with the new input 8

20. freckles

simplify and then you are done

21. monkey*

So, g(8)=x-4/x?

22. freckles

you have to replace all the x's with 8 not just one of the x's

23. freckles

whatever you replace the first x with you have to replace the other x's on the other side also

24. monkey*

g(8)=8-4/8?

25. freckles

yes

26. monkey*

So simplify 8-4/8?

27. monkey*

@freckles

28. freckles

yes

29. freckles

I don't think teachers like unsimplified answers :p

30. monkey*

31. monkey*

And no, they don't lol.

32. monkey*

Thank you so much for the help!

33. freckles

for me to check your answer I have to know one this is it $g(x)=\frac{x-4}{x} \text{ or } g(x)=x-\frac{4}{x}$

34. monkey*

First one.

35. freckles

so it is actually g(x)=(x-4)/x not g(x)=x-4/x so g(8)=(8-4)/8=4/8=1/2 or .5

36. freckles

$f(x)=x^2+7 \\ g(x)=\frac{x-4}{x} \\ g(f(-1)) = g((-1)^2+7)=g(1+7)=g(8)=\frac{8-4}{8}=\frac{4}{8}=\frac{1}{2}$

37. monkey*

So what would be the answer?

38. freckles

well we just got that g(f(-1))=1/2 and we were looking for the value of g(f(-1)) so we are done

39. monkey*

So .5 would be the answer?

40. freckles

yes a half is the answer

41. freckles

1/2 or .5 you choose the form or write in the desired form your teacher wants

42. freckles

I like pretty fractions so I would write 1/2 :p

43. monkey*

Okay, thanks. And she wants a fraction. Could you help with one more?

44. freckles

sure i can try

45. monkey*

Determine the standard form of the equation of the line that passes through (-7,8) and (0,-8).

46. monkey*

Determine the standard form of the equation of the line that passes through (-7,8) and (0,-8).

47. freckles

point slope form for a line looks like: $y-y_1=m(x-x_1) \\ \text{ where } m \text{ is the slope of the line } \\ \text{ and } (x_1,y_1) \text{ is a point on the line }$

48. freckles

you might want to go ahead and find m first

49. freckles

if you have two points you can find m by using the following formula: $m=\frac{y_2-y_1}{x_2-x_1}$

50. freckles

we do have two points so we can use that formula

51. monkey*

M=-8-8/0-(-7)?

52. freckles

$(x_1,y_1) \text{ and } (x_2,y_2) \\ (-7,8) \text{ and } (0,-8) \\ \text{ so replace } x_1 \text{ with } -7 \\ \text{ and replace } y_1 \text{ with } 8 \\ \text{ replace } x_2 \text{ with } 0 \\ \text{ finally replace } y_2 \text{ with } -8 \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{-8-8}{0-(-7)}$ yes if you mean m=(-8-8)/(0-(-7))

53. freckles

and yes you simplify this value for m that we just found

54. freckles

so -8-8=-(8+8)=-16 right and what about the bottom 0-(-7)=-(-7)=...

55. monkey*

M=-16/7 ?

56. freckles

bingo! m is -16/7

57. freckles

$y-y_1=m(x-x_1) \\ \text{ where } m \text{ is the slope of the line } \\ \text{ and } (x_1,y_1) \text{ is a point on the line } \\$ $(x_1,y_1) \text{ and } (x_2,y_2) \\ (-7,8) \text{ and } (0,-8) \\ \text{ so replace } x_1 \text{ with } -7 \\ \text{ and replace } y_1 \text{ with } 8 \\ \text{ replace } x_2 \text{ with } 0 \\ \text{ finally replace } y_2 \text{ with } -8 \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{-8-8}{0-(-7)}$ so let's go back to our point slope form for a line... $y-y_1=m(x-x_1) \\ \text{ we just found } m \text{ to be } \frac{-16}{7} \\ \\ \text{ replace } m \text{ with } \frac{-16}{7}$ $y-y_1=\frac{-16}{7}(x-x_1)$

58. freckles

now really the last thing to do is put in x1 and y1

59. freckles

you can use the value I called x1 and y1 above

60. monkey*

Okay, soo?

61. freckles

that is all that is really left is to replace x1 and y1 have you done that?

62. monkey*

63. freckles

" $y-y_1=m(x-x_1) \\ \text{ where } m \text{ is the slope of the line } \\ \text{ and } (x_1,y_1) \text{ is a point on the line } \\$ $(x_1,y_1) \text{ and } (x_2,y_2) \\ (-7,8) \text{ and } (0,-8) \\ \text{ so replace } x_1 \text{ with } -7 \\ \text{ and replace } y_1 \text{ with } 8 \\ \text{ replace } x_2 \text{ with } 0 \\ \text{ finally replace } y_2 \text{ with } -8 \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{-8-8}{0-(-7)}$ so let's go back to our point slope form for a line... $y-y_1=m(x-x_1) \\ \text{ we just found } m \text{ to be } \frac{-16}{7} \\ \\ \text{ replace } m \text{ with } \frac{-16}{7}$ $y-y_1=\frac{-16}{7}(x-x_1)$ " You should see somehwere here that I called a number x1 and I called a number y1...

64. freckles

remember the point (-7,8) ?

65. freckles

x1 was -7 y1 was 8?

66. monkey*

Yes! I feel stupid now.

67. freckles

remember the last thing really is just to replace x1 with -7 and y1 with 8..

68. freckles

thought this form we are looking at is point-slope form

69. freckles

70. freckles

ax+by=c <--so this is the form we want

71. monkey*

So, -7x+8y=-16/7 ?

72. freckles

your line should be $y-8=\frac{-16}{7}(x-(-7)) \\ \text{ or you could have actually used the point } (x_2,y_2) \text{ instead } \\ y-(-8)=\frac{-16}{7}(x-0) \\ \text{ doesn't matter you just have to plug in a point that is on the line for } \\ (x_1,y_1)$ so either one we play with will give us the same answer... you pick the first one or second one... I honestly think the second one looks prettier

73. freckles

$y+8=\frac{-16}{7}x \\ \text{ note: all I did here was do } -(-8)=8 \text{ and } x-0=x \\ \text{ anyways we have so far } y+8=\frac{-16}{7}x \\ \text{ we want our } x \text{ terms and } y \text{ terms on the same side } \\ \text{ so subtract } y \text{ on both sides } 8=\frac{-16}{7}x-y \\$ in some people's eyes this may be done by some prefer the number in front of x to be positive so you you can multiply both sides by -1 $-8=\frac{16}{7}x+y$ now some prefer just to also have integers coefficients and 16/7 is definitely not an integer so you could also multiply both sides by 7 $-56=16x+7y$

74. freckles

this is probably the most secure way to write it because it fits everyones' expectation of standard form for a line

75. freckles

Ax+By=C A is positive integer B is integer C is integer I think this is the most strict definition of a line in standard form

76. monkey*

I have another one just like this except the points are (9,-4) and (6,4). I do everything the same way, right?

77. freckles

yep

78. freckles

put in point slope form then put in standard form

79. monkey*

Okay, thank you.

80. freckles

np