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anonymous

  • one year ago

Which of these shows the graph for the quadratic function y = x^2 + 2x + 3?

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @ganeshie8

  3. anonymous
    • one year ago
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    @Ashleyisakitty

  4. anonymous
    • one year ago
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    @BlossomCake

  5. anonymous
    • one year ago
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    @Nnesha

  6. anonymous
    • one year ago
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    @UsukiDoll

  7. anonymous
    • one year ago
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    @Misty1212

  8. anonymous
    • one year ago
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    @Michele_Laino

  9. UsukiDoll
    • one year ago
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    ok first of all do not spam and mass tag. I wouldn't want you to end up getting a warning or suspended as this is against OpenStudy's Code of Conduct and Terms of Service.

  10. anonymous
    • one year ago
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    hey your picture is rolanda!

  11. UsukiDoll
    • one year ago
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    Everyone, I got this question under control, so anyway... we need to create a chart that looks like x y = x^2+2x+3 then we just select any x... and solve for that y.... once we have done that, we can pick more x's that will give more y = results and we will have more coordinate points.. For example, if my x = 0 Then x y = x^2+2x+3 0 y=0^2+2(0) +3 0 y =3 so or first coordinate point is (0,3). You graph that on your scratch paper... then you let x be any positive or negative number and continue until your graph matches one of the choices you are given.

  12. anonymous
    • one year ago
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    no please stay i really need help

  13. UsukiDoll
    • one year ago
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    but there are mods in here and I'm not going to give direct answers... my task as a tutor is to teach and guide, not spoonfeed.

  14. anonymous
    • one year ago
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    they are here to help too

  15. UsukiDoll
    • one year ago
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    I already gave you an example on how to solve for this ... please try to understand and attempt the work... then you can answer here and I'll check to see if you're on the right track or not.

  16. anonymous
    • one year ago
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    i dont understand @UsukiDoll explanation is it possible for someone else to explain it like @ganeshie8 @Michele_Laino

  17. UsukiDoll
    • one year ago
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    @ganeshie8 @Michele_Laino... did I not just give an example and explain this thoroughly?

  18. anonymous
    • one year ago
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    u did but im just not understanding it

  19. Michele_Laino
    • one year ago
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    I think you gave a good explanation @UsukiDoll :)

  20. UsukiDoll
    • one year ago
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    see....I'm being thrown under the bus here.

  21. anonymous
    • one year ago
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    im not throwing u under the bus... i have a disease where i need everything to be explained extra cause i cant remember concepts and my brain is messed up cause it hurts to think

  22. anonymous
    • one year ago
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    its a birth defect... :(

  23. anonymous
    • one year ago
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    its called Cerebellar Ataxia

  24. anonymous
    • one year ago
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    @ganeshie8

  25. Michele_Laino
    • one year ago
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    another way is: the x-ccordinate of the vertex of a parabola like this: \[y = a{x^2} + bx + c\] is: \[x = - \frac{b}{{2a}}\] now, by comparison with your parabola, what are the coefficient a, b, and c, and what is the x-coordinate of the vertex of your parabola?

  26. anonymous
    • one year ago
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    whats a parabloa

  27. UsukiDoll
    • one year ago
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    there is more than one way to solve this...maybe that method is too complex for OP?

  28. anonymous
    • one year ago
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    pleae help me:(

  29. UsukiDoll
    • one year ago
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    alright.. I'll try again... what she gave was from College Algebra and seeing as you're not at level yet, we'll go back to what I've just put earlier.

  30. UsukiDoll
    • one year ago
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    we are given \[y = x^2 + 2x + 3\] and the question asks which graph is the same as the quadratic equation

  31. Michele_Laino
    • one year ago
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    I'm sorry, please ignore my hint @barbdewysveck1

  32. anonymous
    • one year ago
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    please help me :(

  33. UsukiDoll
    • one year ago
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    what we can do....is to create a table a table like x y =x^2+2x+3

  34. UsukiDoll
    • one year ago
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    so, we can pick any x and plug it in the y=x^2+2x+3

  35. UsukiDoll
    • one year ago
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    For example, if I pick x = 0. I replace all x's in the y equation x y=x^2+2x+3 0 y=(0)^2+2(0)+3 0^2, also means 0 written twice (0)(0) = 0 2(0) means 2 x 0 which is also 0 so now we have 0 0+0+3 (we add this together and that's 3) so you're first coordinate point is (0,3)

  36. anonymous
    • one year ago
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    so b is the answer

  37. UsukiDoll
    • one year ago
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    well... we only have one point... we need more than (0,3) for a clearer picture

  38. UsukiDoll
    • one year ago
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    so we continue by picking another x... how about if x = 1 so I plug x = 1 into y=x^2+2x+3 1 y=(1)^2+2(1)+3 (1)^2 means (1)(1) = 1 x 1 = 1 2(1) means 2 x 1 and that's 2 so we add up 1+2+3 and we have y = 6 so our next point is (1,6)

  39. UsukiDoll
    • one year ago
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    |dw:1434428975864:dw|

  40. UsukiDoll
    • one year ago
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    I connect those two points after I graph them |dw:1434429072437:dw|

  41. anonymous
    • one year ago
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    its A!!!!!!!!!

  42. UsukiDoll
    • one year ago
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    yes it is!

  43. anonymous
    • one year ago
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    yay can u help me with more

  44. UsukiDoll
    • one year ago
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    because we started at (0,3) and by plotting (1,6) we noticed that the graph was really high up... so only 1 graph was the same as our drawing

  45. UsukiDoll
    • one year ago
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    sure

  46. anonymous
    • one year ago
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  47. UsukiDoll
    • one year ago
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    ok... all we need to is to draw out y=x^2+2x-3 using the same techniques we did and from there we can find the vertex. The vertex of a parabola is the point where the parabola crosses its axis of symmetry.

  48. UsukiDoll
    • one year ago
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    we can just pick our x to be -1,0, and 1 x y=x^2+2x-3 0 y=0^2+2(0)-3 as before 0^2 means writing 0 twice (two times ) but 0 x 0 = 0 similarly 2(0) means 2 x 0 which is again 0 so we have 0+0-3 which is -3 our first point is (0,-3)

  49. UsukiDoll
    • one year ago
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    |dw:1434429432668:dw|

  50. UsukiDoll
    • one year ago
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    it's not drawn to scale... so the graph is a bit ugly...it's just a rough sketch

  51. anonymous
    • one year ago
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    so the answer is B!!!!!!!!

  52. UsukiDoll
    • one year ago
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    umm not yet.. for many reasons 1. we don't have enough information 2. we aren't sure if the graph is really above or below the x-axis

  53. UsukiDoll
    • one year ago
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    we are also not sure if it's on the x-axis or y-axis.

  54. anonymous
    • one year ago
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    my teacher said to look at this

  55. UsukiDoll
    • one year ago
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    x y=x^2+2x-3 so if x = 1 y = (1)^2+2(1)-3 = 1+2-3=3-3=0 so the next point is (1,0) if x = -1 y=(-1)^2+2(-1)-3 = 1-2-3 =-1-3=-4 (-1,-6)|dw:1434429719251:dw|

  56. anonymous
    • one year ago
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    since the example she gave me is +3 and mine is -3 im just gonna go with the opposite of the answer

  57. anonymous
    • one year ago
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    so im gonna go with b

  58. UsukiDoll
    • one year ago
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    I don't think that's a great idea. . . in fact.... that's not a good strategy

  59. UsukiDoll
    • one year ago
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    there is more than one method to solve these problems as long as it doesn't break any math rules... we are using the graphing method to find the vertex which is the lowest or highest point of the graph. In this case, it's the lowest point of the graph which is caused by (-1,-4)

  60. UsukiDoll
    • one year ago
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    |dw:1434430022195:dw|

  61. anonymous
    • one year ago
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    so it is b

  62. UsukiDoll
    • one year ago
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    I need another point to complete this parabola... if I have x = -3 I should have y = 0 I have y = 0 in the right, now I need y =0 in the left y=x^2+2x-3 = (-3)^2+2(-3)-3 = 9-6-3 =3-3=0 |dw:1434430081244:dw|

  63. anonymous
    • one year ago
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    the answer is b

  64. UsukiDoll
    • one year ago
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    |dw:1434430156778:dw| yes ... do you know why?

  65. anonymous
    • one year ago
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    yep

  66. UsukiDoll
    • one year ago
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    ok, can you explain to me why the answer is below the x -axis ?

  67. anonymous
    • one year ago
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    ok this next one i already answered i just need u to say if im right or not, k?

  68. UsukiDoll
    • one year ago
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    alright but can you please answer my question first? I want to see if you understand why it's below the x-axis

  69. anonymous
    • one year ago
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    because the vertex of the parabola is the point where the parabola crossed its axis of symmetry and that happened under the x-axis. BOOM!!!!!!!!!

  70. UsukiDoll
    • one year ago
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    yes and also (-1,-4) is the lowest point in the graph... and since the vertex is (-1,-4) it is indeed under the x-axis.. ok we can move on to the next question .

  71. anonymous
    • one year ago
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    i think the answer is 1

  72. UsukiDoll
    • one year ago
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    ok this time we have to use the discriminant which is b^2-4ac our equation y=-4x^2+3x+2 is in the form of y=ax^2+bx+c. So we let a = -4, b =3, and c = 2 plugging these values into the discriminant gives us (3)^2-4(-4)(2) 9-4(-8) = 9+32=41 I got a massive number.. I'm reading that if the discriminant is 0, there is 1 solution discriminant is negative -> no solution discriminant is positive -> 2 solutions

  73. anonymous
    • one year ago
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    so the answer is 2?

  74. UsukiDoll
    • one year ago
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    I need the quadratic equation to go further.. hold on \[\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 \pm \sqrt {41} }}{{-8}}} \\ \end{array}\]

  75. anonymous
    • one year ago
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    kk

  76. UsukiDoll
    • one year ago
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    \[\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 + \sqrt {41} }}{{-8}}} \\ \end{array}\] \[\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 - \sqrt {41} }}{{-8}}} \\ \end{array}\]

  77. UsukiDoll
    • one year ago
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    the negatives cancel out for the second one \[\LARGE \begin{array}{*{20}c} {x = \frac{{ 3 + \sqrt {41} }}{{8}}} \\ \end{array}\]

  78. UsukiDoll
    • one year ago
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    \[\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 + \sqrt {41} }}{{-8}}} \\ \end{array}\] \[\LARGE x = \frac{3}{8}-\frac{\sqrt{41}}{8}\]

  79. UsukiDoll
    • one year ago
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    the discriminant was positive to begin with... and I don't see any negative signs inside the radical, so we have 2 solutions

  80. anonymous
    • one year ago
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    thx

  81. anonymous
    • one year ago
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    can u check this one too?

  82. anonymous
    • one year ago
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    @UsukiDoll

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