## anonymous one year ago The vertex of x2 + 2x - 4y + 9 = 0 A. (-1,2) B. (1,2) C. (-1,-3) D. (1,-3)

1. anonymous

Im not really sure how to do this problem

2. anonymous

-b/2a

3. UsukiDoll

it's one of those shortcut formulas When working with the vertex form of a quadratic equation $h = \frac{-b}{2a}$ and k=f(h) . The a and b portions come from $ax^2+bx+c$

4. anonymous
5. anonymous

Ok look first let us get the y to a side alone so you get: y=x^2/4 + x/2 + 9/2 right?

6. anonymous

so plus 4y?

7. anonymous

The to get the vertex we should find the derivative of y and equate it to 0 correct?

8. UsukiDoll

derivatives have nothing to do with finding the vertex.

9. UsukiDoll

derivatives is in all levels of Calculus... we're just dealing with quadratic equations... which is College Algebra or even lower.

10. UsukiDoll

get this equation in this form $y=ax^2+bx+c$

11. anonymous

4y=x^2+2x+9

12. anonymous

like that?

13. UsukiDoll

yeah let me double check this for a minute.

14. UsukiDoll

$x^2 + 2x - 4y + 9 = 0 \rightarrow x^2+2x+9=4y$ because we are sort of dealing with this backwards, but I don't think it should be a problem

15. UsukiDoll

now divide the entire equation by 4

16. anonymous

Oh really smartipants let me enlighten you the derivative is the slope of the tan at a specific pt so when you find the derivative of a function and you equate to 0 that means your dealing with the line parallel to the x -axis (slope=0) which is the vertex hahahaha everything works on derivates you clever one I dont advise to tell anybody this again because they'll laugh at you :)

17. anonymous

@UsukiDoll

18. UsukiDoll

And I just reported you for that comment ^

19. anonymous

Dylan are you familiar with derivates or not? So ill continue explaining?

20. UsukiDoll

and if OP isn't familiar with derivatives, that technique shouldn't be used.

21. anonymous

i am not familiar with it sorry..

22. UsukiDoll

See I told you that's a bad idea @joyraheb. And you trying to outsmart me makes it even worse. No one should trust anyone with a lower SS.

23. anonymous

Hahhahahah report thats your way to defend your mathematics 😂 did I even curse well I'll give you an advise, beware of your false knowledge; it is more dangerous than ignorance. :)

24. UsukiDoll

Being rude on here counts as getting you reported man... and Dylan already admit to not knowing derivatives, so that technique you are suggesting isn't going to work.

25. anonymous

y=(x^2/4)+.5+2.25

26. anonymous

is that is divided by 4?

27. UsukiDoll

keep your equation in fraction form. Decimals are messy.

28. UsukiDoll

$x^2+2x+9=4y \rightarrow \frac{x^2}{4}+\frac{1}{2}x+\frac{9}{4}$

29. anonymous

okay i see

30. UsukiDoll

$\frac{1}{4}x^2+\frac{1}{2}x+\frac{9}{4} = y$

31. Owlcoffee

Any parabola can be represented by the parallelism they posses on either axis of reference (xoy). We can take the expression you posted to the structure: $y=ax^2+bx+c$ This is also the equation that represents a parabola parallel to the y-axis, so I will, without proof give you the structure of the coordinates that represent the vertice: $V(-\frac{ b }{ 2a },\frac{ 4ac-b^2 }{ 4a })$ So, taking the conic equation you gave: $x^2+2x-4y+9=0$ We will isolate the "y" term in order to make the coefficients "a", "b" and "c" evident: $y=\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 }$ And then you use the structure I gave you, to find the coordinates of the vertex.

32. UsukiDoll

so far so good everyone.

33. UsukiDoll

so now we let $a= \frac{1}{4}, b =\frac{1}{2}, c =\frac{9}{4}$ and plug it into $V(-\frac{ b }{ 2a },\frac{ 4ac-b^2 }{ 4a })$

34. anonymous

how can i plug it in if i don't know what a is?

35. UsukiDoll

remember our quadratic equation . here is the standard form $ax^2+bx+c$ and you have this equation $y=\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 }$ so comparing these two equations we have our a b and c

36. anonymous

but you just said a = the equation and i don't know what a is so i can't plug it in the vertex formula

37. Owlcoffee

The coefficient of "x^2"

38. UsukiDoll

please read my previous comment. you can get a, b, and c, by comparing your equation to the standard. you will see that a = 1/4, b =1/2, and c=9/4

39. anonymous

okay

40. UsukiDoll

$ax^2+bx+c$ <- standard form $\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 }$ < - your equation.

41. UsukiDoll

so a = 1/4, b =1/2, and c=9/4

42. UsukiDoll

our vertex formula $V(-\frac{ b }{ 2a },\frac{ 4ac-b^2 }{ 4a })$ you just have to plug a, b, and c into these formulas, then we have our vertex.

43. UsukiDoll

let's do the first one together if b = 1/2 and a = 1/4 $\frac{-b}{2a}$ so we place b =1/2 and a =1/4 in that formula $\LARGE \frac{-\frac{1}{2}}{2(\frac{1}{4})}$ $\LARGE \frac{-\frac{1}{2}}{(\frac{2}{4})}$

44. anonymous

i got (2,1)

45. anonymous

wait hold on

46. anonymous

(1,?) i almost got it

47. UsukiDoll

well let's go a bit slowly on this . $\LARGE \frac{-\frac{1}{2}}{(\frac{2}{4})}$ the bottom fraction can be reduced further if we divide 2 on the numerator and denominator on 2/4 we have 1/2 $\LARGE \frac{-\frac{1}{2}}{(\frac{1}{2})}$ then flip the second fraction and we should have $\frac{-1}{2} \times \frac{2}{1}$

48. anonymous

(1,2)???

49. anonymous

is that right?

50. anonymous

i just got confused on plugging it in but then i did it what i think was right and thats the answer i got

51. UsukiDoll

what's $\frac{-1}{2} \times \frac{2}{1}$ what's -1 x 2 and what's 2 x 1?

52. anonymous

-2 and 2

53. UsukiDoll

yes so we have $\frac{-2}{2}$ now what's -2 divided by 2

54. anonymous

-1

55. UsukiDoll

yes that's the first part of our vertex.

56. anonymous

(-1,?)

57. UsukiDoll

$V(-1,\frac{ 4ac-b^2 }{ 4a })$ a=1/4, b =1/2, c =9/4

58. anonymous

2

59. anonymous

(-1,2)

60. UsukiDoll

we're lucky because the denominator will just be 1 so we just have the numerator to deal with $\frac{4}{4} \rightarrow 1$

61. UsukiDoll

$4ac-b^2$ is all we need to deal with $\[4\frac{1}{4}\frac{9}{4}-(\frac{1}{2})^2$ \]

62. UsukiDoll

so let's deal with the right part of this formula. what's $(\frac{1}{2})^2 \rightarrow \frac{1}{2} \times \frac{1}{2}$

63. anonymous

1/4

64. UsukiDoll

yeah. $\[4\frac{1}{4}\frac{9}{4}-\frac{1}{4}$

65. UsukiDoll

so now what's $4 \times \frac{1}{4} \times \frac{9}{4}$ the fraction parts would be easier to do first

66. anonymous

its 2.25

67. anonymous

then minus the .25

68. anonymous

it equals 2

69. UsukiDoll

70. anonymous

decimals make more sense to me

71. anonymous

(-1,2)

72. UsukiDoll

fractions are easier to deal with though...

73. UsukiDoll

$4 \times \frac{1}{4} \times \frac{9}{4} -\frac{1}{4}$ $4 \times \frac{9}{16} -\frac{1}{4}$ $\frac{36}{16} -\frac{1}{4}$ reducing the left fraction by dividing 4 on the numerator and denominator $\frac{9}{4} -\frac{1}{4}$ same numerator so we have 8/4 which is 2 :)

74. UsukiDoll

yay we got our vertex!

75. anonymous

thank you!