## anonymous one year ago Help, will fan and medal. need someone to teach me please!

1. anonymous

What is the equation of the following graph?

2. anonymous

@LynFran can you help me with this? please

3. anonymous

@Robert136 can you help?

4. anonymous

you know the vertex so that makes it easier

5. anonymous

i dont know the vertex...i dont get these at all.

6. anonymous

owait do you need the quadractic or both the quadratic and the dotted line...

7. anonymous

idk..its something to do with ellipses, parabola, or hyperbola...that what the section was about

8. anonymous

i can help with the parabola, not sure about that line though

9. Loser66

What is the directrix? y=??

10. Loser66

What is the vertex? You need both them to write out the equation. Please, answer.

11. anonymous

the vertex is (1,4)

12. anonymous

whats the directrix.... i dont think ive learned those

13. Loser66

The red line below to the graph is a directrix

14. Loser66

if you don't know that method, ok, pick 3 points on the graph. Please.

15. anonymous

y = 3

16. anonymous

thats the directrix

17. Loser66

Now, what method do you want me to follow?? directrix method or standard one?

18. anonymous

which one do u think would be the easiest to understand?

19. Loser66

Depend on you. I am ok with both

20. Loser66

Standard one, ok?

21. anonymous

okay

22. Loser66

what is the standard form of a parabola?

23. anonymous

y=ax^2+bx+c

24. Loser66

YYYYYYYYYYYYYES

25. Loser66

now, 3 points Vertex (1,4) you pick other two.

26. anonymous

just 2 random points?

27. Loser66

Yes, but for convenience, you should pick the point whose coordinates are integers like (-1,5). Got what I meant?

28. anonymous

so the points where the red line cross over?

29. Loser66

no, the point on the parabola

30. anonymous

like (3,5)

31. Loser66

Yes,

32. Loser66

Now, starting Vertex \((\color{red}{1}, \color{blue}{4})\) Plug them into the standard form like \(\color{blue}{4}= a\color{red}{1^2}+b\color{red}{1}+c\) you have 4 = a +b +c, right? name it as (*) Do the same with other two point, what do you get?

33. anonymous

im lost.. im sory. my computer froze sorry about the late anwser

34. anonymous

@Loser66 ?

35. anonymous

5=a-b+c 5=5a+5b+c

36. anonymous

@Loser66 am I understanding it?

37. anonymous

@Mehek14

38. Loser66

5=a -b +c, yes but for (3,5) it is not right 5 = a(3^2) +b(3) +c , that is 5= 9a +3b+c

39. Loser66

Now, you have 3 equations, 3 unknowns a, b, c. You can solve for them a + b + c = 4 (*) a - b + c = 5 (**) 9a + 3b + c = 5 (***)

40. Loser66

Take (*) -(**) , what do you get?

41. anonymous

-1

42. anonymous

@Loser66

43. anonymous

so i have -1 left, and 9a + 3b + c = 5

44. Loser66

|dw:1439603548746:dw|

45. Loser66

Take (***) -(**)

46. anonymous

8a + 4b

47. Loser66

|dw:1439603653405:dw|

48. Loser66

Replace a = 1/4, b = -1/2 in (*) you get c = ??

49. anonymous

c=4.25

50. Loser66

yes, or 17/4 Now, plug back to standard form, y = ax^2 +bx+c . that is your answer.

51. anonymous

y=ax^2+bc+17/4?

52. Loser66

a = 1/4 b= -1/2 c = 17/4 Plug in \(y= ax^2+bx +c\)

53. anonymous

okay thank you. can i tag u if i have anymore questions? i like tht ur showing me instad of telling me

54. misty1212

HI!!

55. misty1212

there might be a somewhat easier way to get this, using two points although above method certainly works

56. anonymous

thank you. im sure ill run into more of these, do u mind if i tag u if i need help?