Help, will fan and medal. need someone to teach me please!

- anonymous

Help, will fan and medal. need someone to teach me please!

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- anonymous

What is the equation of the following graph?

##### 1 Attachment

- anonymous

@LynFran can you help me with this? please

- anonymous

@Robert136 can you help?

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## More answers

- anonymous

you know the vertex so that makes it easier

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- Loser66

What is the directrix? y=??

- Loser66

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

- anonymous

the vertex is (1,4)

- anonymous

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

- Loser66

The red line below to the graph is a directrix

- Loser66

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

- anonymous

y = 3

- anonymous

thats the directrix

- Loser66

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

- anonymous

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

- Loser66

Depend on you. I am ok with both

- Loser66

Standard one, ok?

- anonymous

okay

- Loser66

what is the standard form of a parabola?

- anonymous

y=ax^2+bx+c

- Loser66

YYYYYYYYYYYYYES

- Loser66

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

- anonymous

just 2 random points?

- Loser66

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

- anonymous

so the points where the red line cross over?

- Loser66

no, the point on the parabola

- anonymous

like (3,5)

- Loser66

Yes,

- 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?

- anonymous

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

- anonymous

@Loser66 ?

- anonymous

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

- anonymous

@Loser66 am I understanding it?

- anonymous

@Mehek14

- 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

- 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 (***)

- Loser66

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

- anonymous

-1

- anonymous

@Loser66

- anonymous

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

- Loser66

|dw:1439603548746:dw|

- Loser66

Take (***) -(**)

- anonymous

8a + 4b

- Loser66

|dw:1439603653405:dw|

- Loser66

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

- anonymous

c=4.25

- Loser66

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

- anonymous

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

- Loser66

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

- anonymous

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

- misty1212

HI!!

- misty1212

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

- anonymous

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

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