anonymous one year ago For the given quadratic equation convert into vertex form, find the vertex, and find the value for x = 6. y= -2x^2 + 2x +2?

1. anonymous

@IrishBoy123

2. anonymous

@Michele_Laino

3. Michele_Laino

if we make this substitution: x=6 into your quadratic equation, we get: $\Large \begin{gathered} y = - 2 \times {6^2} + 2 \times 6 + 2 = \hfill \\ \hfill \\ = - 2 \times 36 + 12 + 2 = ...? \hfill \\ \end{gathered}$

4. anonymous

-58

5. Michele_Laino

that's right!

6. anonymous

So that's the value of x, correct?

7. Michele_Laino

that's the value of y, when x=6

8. anonymous

So how do we find the value of x and find the vertex?

9. Michele_Laino

our task was to find the value of y, when x=6

10. Michele_Laino

the general formula of a parabola is: $\Large y = a{x^2} + bx + c$

11. anonymous

So there is nothing else to the problem it's done that simple lol

12. Michele_Laino

we have to find the coordinates of the vertex, and the vertex form of the equation of your parabola

13. Michele_Laino

so, as I wrote before, the general equation of a parabola, whose axis is parallel to y axis, is: $\Large y = a{x^2} + bx + c$ now, please by comparison with the equationof your parabola, what are the coefficients a, b, and c?

14. anonymous

6 and -58?

15. Michele_Laino

hint: |dw:1435870805196:dw|

16. anonymous

ok so its 2,2,and 2

17. Michele_Laino

better is a=-2

18. anonymous

A=-2, b=2,c=2

19. Michele_Laino

that's right!

20. Michele_Laino

now the x-coordinate of the vertex is: $\Large x = - \frac{b}{{2a}} = ...?$

21. Michele_Laino

22. Michele_Laino

that*

23. anonymous

X=$-\frac{ 2b }{ 2a}$

24. Michele_Laino

hint: $\Large x = - \frac{b}{{2a}} = - \frac{2}{{2 \times \left( { - 2} \right)}} = ...?$

25. anonymous

I really dont know

26. Michele_Laino

why?

27. Michele_Laino

$\Large x = - \frac{b}{{2a}} = - \frac{2}{{2 \times \left( { - 2} \right)}} = - \frac{2}{{ - 4}} = ...?$

28. anonymous

I Oh ok i thought it was like a trick next step lol

29. anonymous

$\frac{ 1 }{ 2}$

30. anonymous

negative

31. Michele_Laino

that's right!

32. Michele_Laino

no, it is 1/2

33. anonymous

oh ok

34. Michele_Laino

next, the y-coordinate of the vertex is given by the subsequent formula: $\Large y = - \frac{{{b^2} - 4ac}}{{4a}} = ...?$ as before, please substitute your coefficients into that formula

35. anonymous

$y=\frac{ 2^{2}-4ac }{ 4a}$

36. Michele_Laino

hint: $\Large y = - \frac{{{b^2} - 4ac}}{{4a}} = - \frac{{{2^2} - \left\{ {4 \times \left( { - 2} \right) \times 2} \right\}}}{{4 \times \left( { - 2} \right)}} = ...?$

37. anonymous

$\frac{ 20 }{ -8}$

38. Michele_Laino

I think: 20/8

39. Michele_Laino

am I right?

40. anonymous

yes

41. anonymous

so its $2\frac{ 1 }{ 2 }$

42. Michele_Laino

ok! so the vertex V of our parabola is this point: $\Large V = \left( {\frac{1}{2},\frac{5}{2}} \right)$ since: $\Large \frac{{20}}{8} = \frac{5}{2}$

43. anonymous

So my answers are -58 for x inputting into the y and my vertex is (1/2) (5/2)? correct

44. Michele_Laino

yes! correct!

45. anonymous

yay we did it lol

46. Michele_Laino

finally, we have to write the vertex form of the equation of your parabola

47. anonymous

Thank you very much for your patience

48. Michele_Laino

:)

49. anonymous

50. Michele_Laino

first step: we can factor out -2 at the right side of your equation, so we get: $\Large y = - 2\left( {{x^2} - x - 1} \right)$

51. anonymous

so do we input abc or leave it like that

52. Michele_Laino

we have to continue with the second step

53. Michele_Laino

here is the second step: $\Large {x^2} - x - 1 = {x^2} - x + \frac{1}{4} - \frac{1}{4} - 1$

54. Michele_Laino

55. anonymous

so wouldnt it just x^2-x-1

56. Michele_Laino

yes! I cahnge the form of x^2-x-1 using the rules of algebra

57. Michele_Laino

change*

58. Michele_Laino

nevertheless the quantity is the same, I change only its algebraic shape

59. anonymous

$x ^{2}-x-1$

60. Michele_Laino

yes that quantity will be the same, I change its form only

61. Michele_Laino

next I can simplify that expression above as below: $\Large \begin{gathered} {x^2} - x - 1 = {x^2} - x + \frac{1}{4} - \frac{1}{4} - 1 = \hfill \\ \hfill \\ = {x^2} - x + \frac{1}{4} - \frac{5}{4} \hfill \\ \end{gathered}$

62. Michele_Laino

now, the quantity: $\large {x^2} - x + \frac{1}{4}$ is a perfect square, namely it is the square of which binomial?

63. anonymous

i have no clue

64. Michele_Laino

hint: please compute this: $\Large {\left( {x - \frac{1}{2}} \right)^2} = ...?$

65. anonymous

|dw:1435879978188:dw|

66. Michele_Laino

perfect!

67. Michele_Laino

so we can write this: $\large \begin{gathered} {x^2} - x - 1 = {x^2} - x + \frac{1}{4} - \frac{1}{4} - 1 = \hfill \\ \hfill \\ = {x^2} - x + \frac{1}{4} - \frac{5}{4} = {\left( {x - \frac{1}{2}} \right)^2} - \frac{5}{4} \hfill \\ \end{gathered}$

68. Michele_Laino

namely: $\large {x^2} - x - 1 = {\left( {x - \frac{1}{2}} \right)^2} - \frac{5}{4}$

69. Michele_Laino

then I substitute that new expression, into the subsequent formula: $\Large y = - 2\left( {{x^2} - x - 1} \right)$

70. Michele_Laino

and I get: $\Large y = - 2\left\{ {{{\left( {x - \frac{1}{2}} \right)}^2} - \frac{5}{4}} \right\}$

71. anonymous

$y=-2\left( x-\frac{ 1 }{ 2 } \right)^{2}+\frac{ 5 }{ 2 }$

72. Michele_Laino

congratulations!! :)

73. anonymous

Yay lol

74. anonymous

So the vertex is 1/2 and 5/2, that's are equation above and -58 is the input?

75. Michele_Laino

yes!

76. anonymous

yay thank you so much!

77. Michele_Laino

:)