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do you see how y = 3x^2 + 6x + 2 is in the form y = ax^2 + bx + c ?

a = 3 b = 6 c = 2?

yep

now plug the values of a,b into h = -b/(2a)
what is the value of h?

h = -1?

yes

now plug that value in for x to find y
y = 3x^2 + 6x + 2
y = 3(-1)^2 + 6(-1) + 2
y = ???

Is it -1?

correct

so k = -1

we have
a = 3, h = -1 and k = -1
plug those into
y = a(x-h)^2 + k

Does x = 6?

where are you getting x = 6 ?

Oh thought x = 6 from the problem. Dont know what x is.

x is just a variable to be left alone really

what did you get when you plugged those values into y = a(x-h)^2 + k

Is it 11?

don't replace x though

Oh keep x as just x?

yes

So is this right? y - 3(x - (-1)^2 + - 1 . I think I set it up wrong.

I meant y =

if you simplified that, you'd get what?

Does the 3 go with the things inside the ( )

|dw:1442024648925:dw|

|dw:1442024665299:dw|

|dw:1442024695181:dw|

Yeah does the 3 multiply with that?

what does `x-(-1)` simplify to?

3x - (-3)^2 - (-1) ?

if you subtract a negative, what are you really doing?

Making it positive?

yeah adding
so `x-(-1)` is the same as `x+1`

making
\[\Large y=3(x-(-1))^2+(-1)\]
turn into
\[\Large y=3(x+1)^2-1\]

Oh okay I see.

What would be the vertex here?

I am not really sure.

Do you put everything together?

well the general vertex form equation
\[\Large y = a(x-h)^2 + k\]
has the vertex `(h,k)`

(1, -1) ?

well recall that h = -1 and k = -1

Oh thats not the vertex?

no but you're close

I do not get how to find it.

it's (-1,-1)
you start with (h,k) and replace h and k with the values we got earlier

h = -b/(2a) = -1 after plugging in a and b
k = -1 after plugging x = -1 into the original equation

Oh I see where it came from.

Is the vertex (-1, -1) or is that used to find the vertex?

Oh okay and then y = 3(x + 1)^2 - 1 is the equation rewritten?

in the alternate form*

that's vertex form, yes

Oh okay thanks for your help. I am very slow in math as anyone can tell.

you're doing fine