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Janu16

  • one year ago

Below are two different functions, f(x)and g(x). What can be determined about their y-intercepts?

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  1. Janu16
    • one year ago
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    f(x) = x + 4 x g(x) -1 8 1 0 2 -4 @DanJS here it is

  2. DanJS
    • one year ago
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    k

  3. DanJS
    • one year ago
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    the y intercept is the point (x,y) = (0,y) when x is zero it will cross the Y axis, can you picture that

  4. Janu16
    • one year ago
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    um nope

  5. DanJS
    • one year ago
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    |dw:1437444302167:dw|

  6. Janu16
    • one year ago
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    oh i see

  7. DanJS
    • one year ago
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    When x axis line is at zero, you are in the y-axis

  8. DanJS
    • one year ago
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    so what can you think about f(x) for it's y-intercept, what would you do ?

  9. Janu16
    • one year ago
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    so y is 4 right?..

  10. DanJS
    • one year ago
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    right, you set X=0, then you get the y-intercept F(x) = 0 + 4

  11. Janu16
    • one year ago
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    so put dot on 4?

  12. DanJS
    • one year ago
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    yes you can place a point at ( x , y) = (0 , 4)

  13. DanJS
    • one year ago
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    similarly, if you want to graph a line, you need 2 points right? so find now the X-intercept by doing the same thing, this time set Y=0 and figure X

  14. Janu16
    • one year ago
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    are you talking about g(x)?than it would be 1...?

  15. DanJS
    • one year ago
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    f(x) = y = x + 4 0 = x + 4 when y is zero, x is -4 second point (x , y) = ( -4 , 0)

  16. Janu16
    • one year ago
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    oh i see

  17. DanJS
    • one year ago
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    no still f(x), similar question, now you can graph the line, you have 2 points, the X and the Y intercepts

  18. DanJS
    • one year ago
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    |dw:1437444769726:dw|

  19. Janu16
    • one year ago
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    why would the 4 be negative? for o,-4

  20. DanJS
    • one year ago
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    fixed sorry

  21. Janu16
    • one year ago
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    ohok

  22. DanJS
    • one year ago
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    FOr g(x), it looks like it is also a line, constant slope

  23. Janu16
    • one year ago
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    so the other 4 would be negative?i thought it would be positive

  24. Janu16
    • one year ago
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    wasnt the point 4,0?

  25. DanJS
    • one year ago
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    No, when y = f(x) = 0 , you get 0 = x + 4 0 - 4 = x + 4 - 4 x = -4 the x intercept is -4

  26. DanJS
    • one year ago
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    plus the slope is +1, the line must be climbing

  27. Janu16
    • one year ago
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    ohok i get it and what were you saying about g(x)?

  28. DanJS
    • one year ago
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    if you calculate the slope between points 1-2 and 2-3 or any 2 of those points, the slope is the same, it is a line of the form y = m*x+ b

  29. DanJS
    • one year ago
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    a line has the same slope everywhere

  30. DanJS
    • one year ago
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    yep, now you hav e y = mx + b y = -4*x + b pick any point and put in (x,y) and solve b

  31. DanJS
    • one year ago
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    b is the y intercept for that form of a line equation, y = (slope)*x + (y int)

  32. DanJS
    • one year ago
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    see when x = 0, you get y=b, the y intercept

  33. Janu16
    • one year ago
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    so what can be determined about both of their y intercepts

  34. DanJS
    • one year ago
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    since you figured the slope is -4 for g(x) y = -4*x+ b use any point on g, i pick (x , y ) = (1 , 0) ... put into equation 0 = -4*(1) + b what is b - (y intercept) for g?

  35. Janu16
    • one year ago
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    -4

  36. DanJS
    • one year ago
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    |dw:1437445491201:dw|

  37. DanJS
    • one year ago
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    b = 4 thjey both have the same y-intercepts

  38. DanJS
    • one year ago
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    0 = -4 + b add 4 both side b = 4

  39. Janu16
    • one year ago
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    ok they all have same y intercepts?

  40. DanJS
    • one year ago
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    yes, if you don't get something let me know

  41. Janu16
    • one year ago
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    i get it so my answer would be they all have same y intercept. thanks and you know how you helped me with f(g((x) problem i am trying to do mine but cant get the answer. f(x) = 2x + 1 and g(x) = −5x + 2, solve for f(g(x)) when x = 3. dont you start with g(x) first?

  42. Janu16
    • one year ago
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    i did -5 times 3 + 2

  43. Janu16
    • one year ago
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    nvm i got it its -25

  44. DanJS
    • one year ago
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    yeah, remember to figure the inner function first, then use that value to move outwards

  45. Janu16
    • one year ago
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    yup i remember and id you have g(f((x) do you start with f first?

  46. DanJS
    • one year ago
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    yep. W[f(G(x))] G then f then W, keeping track of the values to use for each as you work outwards

  47. Janu16
    • one year ago
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    what is w

  48. DanJS
    • one year ago
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    i just added a third function, same idea, work middle to outside

  49. Janu16
    • one year ago
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    ohoki am just learning about 2 function right now in algebra 2 lol but thxs

  50. DanJS
    • one year ago
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    no prob

  51. Janu16
    • one year ago
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    does this \ mean graph is going negative? like that line

  52. DanJS
    • one year ago
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    that means it has a negative slope, for increasing values of X, the Y value will be less and less

  53. DanJS
    • one year ago
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    as you figured the slope was minus 4

  54. Janu16
    • one year ago
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    ohok and i was doing inverse so how do you find inverse of this f(x) = x2 − 16 like it has only 2 things so how do you inverse that

  55. DanJS
    • one year ago
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    If you see an equation in the form y = mx+b and the m is negative, it will do that

  56. Janu16
    • one year ago
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    ok i see it and would it be x=y2-16 than what do you subtract

  57. DanJS
    • one year ago
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    remember how the inverse reflects over that line with slope 1, y = x

  58. DanJS
    • one year ago
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  59. DanJS
    • one year ago
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    yeah just solve for the new y value for the inverse if you need that

  60. Janu16
    • one year ago
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    yas i see how it reflects but how do you solve for y value like do you subtracvt 16

  61. Janu16
    • one year ago
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    i mean add

  62. DanJS
    • one year ago
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    oh, you gots to get the algebra stuff down,, lol

  63. DanJS
    • one year ago
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    \[x = y^2 - 16\] \[x+16=y^2\]

  64. DanJS
    • one year ago
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    take square root of both sides

  65. Janu16
    • one year ago
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    ±square root x+16?

  66. DanJS
    • one year ago
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    yes that is y or f inverse

  67. DanJS
    • one year ago
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    the + is above the x axis and the - is below, as graphed

  68. Janu16
    • one year ago
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    so that would be the answer right?

  69. DanJS
    • one year ago
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    right

  70. Janu16
    • one year ago
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    thx :p

  71. DanJS
    • one year ago
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    not sure what you have to remember for tests, but just recall inverse functions, switch x and y and solve new y is the inverse

  72. DanJS
    • one year ago
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    if it exists, but it probably will if they ask you

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