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anonymous

  • one year ago

how do i draw the graph for m (x)= x^2 + 4x + 9

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  1. DanJS
    • one year ago
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    do you se the general shape of it by the form , it is already in the general form

  2. DanJS
    • one year ago
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    f(x) = ax^2 + bx + c general quadratic form, parabola

  3. anonymous
    • one year ago
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    how do i find the domain for the graph ? i factorized it and i got (x+2)^2 + 9

  4. DanJS
    • one year ago
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    watch the completing the square there, you did that part right, but the constant 9 is wrong

  5. anonymous
    • one year ago
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    it says that x is a real number... obviously cuz the graph is infinity but it also says that x> a and i have to find the value of a.... how do i find a?

  6. DanJS
    • one year ago
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    given m(x) = x^2 + 4x + 9 Graph the function and find the domain? what is it asking

  7. anonymous
    • one year ago
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    its asking the value of a

  8. anonymous
    • one year ago
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    in which x > a

  9. anonymous
    • one year ago
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    oh and i got the completing square... (x+2)^2 + 5

  10. DanJS
    • one year ago
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    the domain of that is going to be all real numbers for possible values of X

  11. DanJS
    • one year ago
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    It states that x is a real number because you want to exclude complex numbers or roots

  12. anonymous
    • one year ago
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    but how do i draw the graph...

  13. anonymous
    • one year ago
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    where will the intersection for the parabola be

  14. DanJS
    • one year ago
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    |dw:1443165464570:dw|

  15. anonymous
    • one year ago
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    oki i finally got it!!

  16. anonymous
    • one year ago
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    thank you so much

  17. DanJS
    • one year ago
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    I would first find the vertex point, the x coordinate of that will be found by calculating -b / (2*a) = -4/2(1) = -2

  18. DanJS
    • one year ago
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    hah, you sure,, kk

  19. anonymous
    • one year ago
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    I'm sorry i got confused in the vertex point part..

  20. DanJS
    • one year ago
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    after you find the vertex, notice the constant on the x^2 term is positive, meaning it opens upwards, and starting at that vertex, there will be no x-axis intercepts

  21. DanJS
    • one year ago
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    if it was -x^2 it woul dopen down, and you would have two x-intercepts to find

  22. anonymous
    • one year ago
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    is it always found using -b / 2*a??

  23. DanJS
    • one year ago
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    yes, the x coordinate of the vertex, then use that value in the function to get the y coordinate of the point. Do you need to see how that formula was found or no ?

  24. DanJS
    • one year ago
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    x = -b/(2a)

  25. DanJS
    • one year ago
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    you were trying to put the function into the 'vertex form' by completing the square. that is possible too, and you can read the point right from the equation

  26. anonymous
    • one year ago
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    oh i didn't know that before... isn't it usually -b +/- (sqrt.) (b^2 -4ac) / 2a

  27. DanJS
    • one year ago
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    that is the quadratic formula, used in finding the values where the graph crosses the x-axis, or the roots of the quadratic

  28. anonymous
    • one year ago
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    oki so to find the vertex i always use the formula used for x^2 graphs

  29. anonymous
    • one year ago
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    thats -b/ 2a as u stated before

  30. DanJS
    • one year ago
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    notice the stuff under the root term turns out to be a negative number b^2-4ac no roots in the real numbers, cant have a square root of a negative number

  31. anonymous
    • one year ago
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    yep so its ignored?

  32. DanJS
    • one year ago
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    that is why you don't see it cross the x axis

  33. anonymous
    • one year ago
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    oh ok

  34. anonymous
    • one year ago
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    i think i get it now

  35. anonymous
    • one year ago
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    once again thanks for the help :)

  36. DanJS
    • one year ago
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    to graph a parabola, i would find the easy points first vertex x-intercepts if any y-intercepts if any that is probably enough to get the general graph

  37. DanJS
    • one year ago
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    if it is y in terms of x, and opens up or down, the domain is all real numbers if it is x in terms of y, and opens right or left, then the domain will be all x to the right or left of the value x=a, the vertex point x value

  38. DanJS
    • one year ago
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    I think that is what they were talking about the x >a thing, but this one does not open sideways, so no 'a' value

  39. DanJS
    • one year ago
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    \[x = y^2\]

  40. DanJS
    • one year ago
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    The domain of that will be all x values greater than or equal to zero

  41. anonymous
    • one year ago
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    is there a reflection to this graph? m ^-1 (x)

  42. DanJS
    • one year ago
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    yes, that would be the graph resulting in reflecting over the line y = x

  43. DanJS
    • one year ago
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    x = y^2 + 4y + 9 you can re-solve for y if you want, but that is the inverse

  44. anonymous
    • one year ago
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    yeah i know how to resolve it.... but the graph drawing part is annoying

  45. DanJS
    • one year ago
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    |dw:1443166512627:dw|

  46. DanJS
    • one year ago
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    you are basically switching the x and y values in every point

  47. DanJS
    • one year ago
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    the new vertex is (5 , -2)

  48. anonymous
    • one year ago
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    u only switch the vertex?

  49. anonymous
    • one year ago
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    what about the coordinates (0, 9)

  50. DanJS
    • one year ago
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    |dw:1443166710633:dw|

  51. DanJS
    • one year ago
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    the + value on the y^2 term means it opens to the right the domain in that case for the inverse is all values of x such that x is larger or equal to 5

  52. anonymous
    • one year ago
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    ok, so for refection we make the subject of the formula x so the other side has the y term... and as the y is positive we open it to the right side... if it was negative it would be the left?

  53. anonymous
    • one year ago
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    so all reflections are either to the right or left for an parabolic graph?

  54. anonymous
    • one year ago
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    its fine i think i got it... thank u so much :)

  55. DanJS
    • one year ago
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    the inverse function is a reflection over both the x and the y axis, combined it is a reflection over the line y = x

  56. DanJS
    • one year ago
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    swap the variables x and y in the function and then solve for y, that will be the inverse , if it exists

  57. DanJS
    • one year ago
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    |dw:1443167505039:dw|

  58. DanJS
    • one year ago
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    every point on the upwards one has a corresponding point on the sideways inverse, the values are just the switched numbers for x and y

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