ali1029 Group Title Please help! how do you write y=x^2-2x-3 in vertex and intercept form using the quadratic formula? 9 months ago 9 months ago

1. ali1029 Group Title

@ganeshie8 @satellite73 do you guys know how to do this? :o

I think it is y=-2x-3

Oops I meant xD y=-2x+1

4. ali1029 Group Title

is that for vertex form or intercept form? :)

Oops damn I put that in slope-intercept... Geez I suck at math xD, I don't think I can help you on vertex

6. ali1029 Group Title

hahah thanks for trying :)

7. ganeshie8 Group Title

use the quadratic formula and find the x-intercepts first

8. ali1029 Group Title

can you help me figure out how to do that? thats where im stuck :(

9. ganeshie8 Group Title

sure :) u knw the quadratic formula ?

10. ali1029 Group Title

sorta :o im just so confused because we've been using so many different forms!

11. ganeshie8 Group Title

dont wry, this is going to be very easy :) First, we find x-intercepts using the quadratic formula.

12. ganeshie8 Group Title

y=x^2-2x-3 a = ? b = ? c = ?

13. ganeshie8 Group Title

$$\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

14. ganeshie8 Group Title

you need to figure out a, b, c values. and plug them above to get the x-intercepts

15. ali1029 Group Title

so a=1 b=2 c=3? what i dont get is that it wants me to put that equation in intercept and vertex form but those have h & k :o

16. ganeshie8 Group Title

b = -2 we will get to vertex/intercept forms later. first, get the x-intercepts :)

17. ganeshie8 Group Title

put them in the formula above, wat do u get for x = ?

18. ali1029 Group Title

1 becaue 1^2 is 1? let me do that equation

19. ganeshie8 Group Title

a = 1 b = -2 c = -3

20. ganeshie8 Group Title

use above, take ur time :)

21. ganeshie8 Group Title

$$\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}$$

22. ali1029 Group Title

i got 1 ± 4 .... is that close? :o

23. ganeshie8 Group Title

$$\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}$$ $$\large x = \frac{2 \pm \sqrt{4 + 12}}{2}$$ $$\large x = \frac{2 \pm \sqrt{16}}{2}$$

24. ganeshie8 Group Title

thats close, just a minor mistake u did... check once

25. ganeshie8 Group Title

$$\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}$$ $$\large x = \frac{2 \pm \sqrt{4 + 12}}{2}$$ $$\large x = \frac{2 \pm \sqrt{16}}{2}$$ $$\large x = \frac{2 \pm 4}{2}$$ $$\large x = 1 \pm 2$$

26. ali1029 Group Title

ohhh i see what i did wrong! thank you :) so know what do i do?

27. ganeshie8 Group Title

that gives, $$x = 1+2 ~~~ or~~~~ x = 1-2$$ $$x = 3, -1$$

28. ganeshie8 Group Title

we're done wid using quadratic formula. we got the x-intercepts : 3, -1

29. ganeshie8 Group Title

so the intercept form would be : y = (x-3)(x+1)

30. ganeshie8 Group Title

wat about vertex form ? how to get $$h$$, $$k$$ ... any ideas ?

31. ali1029 Group Title

wow that was easier than i thought! thank you :) & no, not too sure :(

32. ganeshie8 Group Title

vertec form is also easy, $$h$$ lies exactly at the middle of both the intercepts :- $$h$$ = $$\large \frac{3 + -1}{2}$$

33. ganeshie8 Group Title

$$h$$ = $$\large \frac{3-1}{2}$$

34. ganeshie8 Group Title

$$h$$ = $$\large \frac{2}{2}$$

35. ali1029 Group Title

so h=1?

36. ganeshie8 Group Title

$$h$$ = $$\large 1$$

37. ali1029 Group Title

yay :D

38. ganeshie8 Group Title

so vertex form wud be : $$y = (x-1)^2 + k$$

39. ganeshie8 Group Title

you still need to find $$k$$

40. ganeshie8 Group Title

:)

41. ali1029 Group Title

okay, so how do we find k now? :)

42. ganeshie8 Group Title

put x = h = 1, in given quadratic. it gives u the $$k$$ value

43. ganeshie8 Group Title

y=x^2-2x-3 put x = 1

44. ganeshie8 Group Title

y = 1^2 - 2(1) - 3 = 1 -2 -3 = ?

45. ali1029 Group Title

-4?

46. ganeshie8 Group Title

yup ! thats ur $$k$$

47. ali1029 Group Title

y=(x−1)^2+4 would be the answer? :)

48. ganeshie8 Group Title

who ate the - sign before 4 ?

49. ali1029 Group Title

i messed up again ._. so it would be y=(x−1)2-4?

50. ganeshie8 Group Title

yes :)

51. ali1029 Group Title

thank you so much!! you're amazing!! :) if it isnt too much too ask could you help me with two more equations similar to that? :)

52. ali1029 Group Title

i need to figure out how to put y-5=-2(x+1)^2 in standard and intercept form & y=(x+2)(x-3) in standard and vertex form :o

53. ganeshie8 Group Title

wats the standard form ?

54. ali1029 Group Title

y=ax^2+bx+c?

55. ganeshie8 Group Title

y = ax^2 + bx + c this is the standard form right ?

56. ganeshie8 Group Title

oh yes... thats right !

57. ganeshie8 Group Title

y-5=-2(x+1)^2 add 5 both sides y = -2(x+1)^2 + 5

58. ganeshie8 Group Title

next, use the formula $$(a+b)^2 = a^2 +b^2 + 2ab$$

59. ganeshie8 Group Title

expand right side,

60. ali1029 Group Title

would it be y=-2x+1

61. ganeshie8 Group Title

y = -2(x+1)^2 + 5 = -2(x^2 + 1^2 + 2x ) + 5 = -2(x^2 + 1 + 2x) + 5

62. ganeshie8 Group Title

simplify :)

63. ali1029 Group Title

would it be -2x^2-4x+3?

64. ganeshie8 Group Title

correct

65. ganeshie8 Group Title

standard form : y = -2x^2-4x+3

66. ali1029 Group Title

yesss :D

67. ganeshie8 Group Title

y-5=-2(x+1)^2 to put this guy in intercept form, u need to find its x-intercepts. so put y = 0 0-5=-2(x+1)^2 5/2 = (x+1)^2 $$\pm \sqrt{5/2} = x+1$$ $$-1 \pm \sqrt{5/2} = x$$

68. ganeshie8 Group Title

those are our x-intercepts

69. ganeshie8 Group Title

can u put t he intercept form ?

70. ali1029 Group Title

the intercept form is y=a(x-p)(x-q)

71. ali1029 Group Title

so if the equation is −1±5/2, how do i solve the fraction part?