## anonymous one year ago Which of these shows the graph for the quadratic function y = x^2 + 2x + 3?

1. anonymous

2. anonymous

@ganeshie8

3. anonymous

@Ashleyisakitty

4. anonymous

@BlossomCake

5. anonymous

@Nnesha

6. anonymous

@UsukiDoll

7. anonymous

@Misty1212

8. anonymous

@Michele_Laino

9. UsukiDoll

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

11. UsukiDoll

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

no please stay i really need help

13. UsukiDoll

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

they are here to help too

15. UsukiDoll

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

i dont understand @UsukiDoll explanation is it possible for someone else to explain it like @ganeshie8 @Michele_Laino

17. UsukiDoll

@ganeshie8 @Michele_Laino... did I not just give an example and explain this thoroughly?

18. anonymous

u did but im just not understanding it

19. Michele_Laino

I think you gave a good explanation @UsukiDoll :)

20. UsukiDoll

see....I'm being thrown under the bus here.

21. anonymous

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

its a birth defect... :(

23. anonymous

its called Cerebellar Ataxia

24. anonymous

@ganeshie8

25. Michele_Laino

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

whats a parabloa

27. UsukiDoll

there is more than one way to solve this...maybe that method is too complex for OP?

28. anonymous

pleae help me:(

29. UsukiDoll

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

we are given $y = x^2 + 2x + 3$ and the question asks which graph is the same as the quadratic equation

31. Michele_Laino

I'm sorry, please ignore my hint @barbdewysveck1

32. anonymous

33. UsukiDoll

what we can do....is to create a table a table like x y =x^2+2x+3

34. UsukiDoll

so, we can pick any x and plug it in the y=x^2+2x+3

35. UsukiDoll

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

37. UsukiDoll

well... we only have one point... we need more than (0,3) for a clearer picture

38. UsukiDoll

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

|dw:1434428975864:dw|

40. UsukiDoll

I connect those two points after I graph them |dw:1434429072437:dw|

41. anonymous

its A!!!!!!!!!

42. UsukiDoll

yes it is!

43. anonymous

yay can u help me with more

44. UsukiDoll

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

sure

46. anonymous

47. UsukiDoll

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

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

|dw:1434429432668:dw|

50. UsukiDoll

it's not drawn to scale... so the graph is a bit ugly...it's just a rough sketch

51. anonymous

52. UsukiDoll

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

we are also not sure if it's on the x-axis or y-axis.

54. anonymous

my teacher said to look at this

55. UsukiDoll

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

since the example she gave me is +3 and mine is -3 im just gonna go with the opposite of the answer

57. anonymous

so im gonna go with b

58. UsukiDoll

I don't think that's a great idea. . . in fact.... that's not a good strategy

59. UsukiDoll

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

|dw:1434430022195:dw|

61. anonymous

so it is b

62. UsukiDoll

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

64. UsukiDoll

|dw:1434430156778:dw| yes ... do you know why?

65. anonymous

yep

66. UsukiDoll

ok, can you explain to me why the answer is below the x -axis ?

67. anonymous

ok this next one i already answered i just need u to say if im right or not, k?

68. UsukiDoll

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

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

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

i think the answer is 1

72. UsukiDoll

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

74. UsukiDoll

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

kk

76. UsukiDoll

$\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 + \sqrt {41} }}{{-8}}} \\ \end{array}$ $\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 - \sqrt {41} }}{{-8}}} \\ \end{array}$

77. UsukiDoll

the negatives cancel out for the second one $\LARGE \begin{array}{*{20}c} {x = \frac{{ 3 + \sqrt {41} }}{{8}}} \\ \end{array}$

78. UsukiDoll

$\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 + \sqrt {41} }}{{-8}}} \\ \end{array}$ $\LARGE x = \frac{3}{8}-\frac{\sqrt{41}}{8}$

79. UsukiDoll

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

thx

81. anonymous

can u check this one too?

82. anonymous

@UsukiDoll