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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?
now plug the values of a,b into h = -b/(2a) what is the value of h?
h = -1?
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?
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?
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 ( )
I think you were asking if 3 goes inside the parenthesis? or were you asking if the 3 was being multiplied with the (x-h)^2 term?
Yeah does the 3 multiply with that?
yes the 3 and \(\Large (x-(-1))^2\) are being multiplied it's like saying `3x` is the same as `3 times x`
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?
that is the vertex as you can see on this graph, it's the lowest point or the min point https://www.desmos.com/calculator/idhk1vx8ar
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