Rewrite the equation y = 3x^2 + 6x + 2 in the alternate form and state the vertex point.

- anonymous

Rewrite the equation y = 3x^2 + 6x + 2 in the alternate form and state the vertex point.

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- schrodinger

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- jim_thompson5910

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

- anonymous

a = 3 b = 6 c = 2?

- jim_thompson5910

yep

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## More answers

- jim_thompson5910

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

- anonymous

h = -1?

- jim_thompson5910

yes

- jim_thompson5910

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

- anonymous

Is it -1?

- jim_thompson5910

correct

- jim_thompson5910

so k = -1

- jim_thompson5910

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

- anonymous

Does x = 6?

- jim_thompson5910

where are you getting x = 6 ?

- anonymous

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

- jim_thompson5910

x is just a variable to be left alone really

- jim_thompson5910

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

- anonymous

Is it 11?

- jim_thompson5910

don't replace x though

- anonymous

Oh keep x as just x?

- jim_thompson5910

yes

- anonymous

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

- anonymous

I meant y =

- jim_thompson5910

if you simplified that, you'd get what?

- anonymous

Does the 3 go with the things inside the ( )

- jim_thompson5910

no

- jim_thompson5910

|dw:1442024648925:dw|

- jim_thompson5910

|dw:1442024665299:dw|

- jim_thompson5910

|dw:1442024695181:dw|

- jim_thompson5910

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?

- anonymous

Yeah does the 3 multiply with that?

- jim_thompson5910

yes the 3 and \(\Large (x-(-1))^2\) are being multiplied
it's like saying `3x` is the same as `3 times x`

- jim_thompson5910

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

- anonymous

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

- jim_thompson5910

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

- anonymous

Making it positive?

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

Oh okay I see.

- jim_thompson5910

What would be the vertex here?

- anonymous

I am not really sure.

- anonymous

Do you put everything together?

- jim_thompson5910

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

- anonymous

(1, -1) ?

- jim_thompson5910

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

- anonymous

Oh thats not the vertex?

- jim_thompson5910

no but you're close

- anonymous

I do not get how to find it.

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

Oh I see where it came from.

- anonymous

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

- jim_thompson5910

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

- anonymous

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

- anonymous

in the alternate form*

- jim_thompson5910

that's vertex form, yes

- anonymous

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

- jim_thompson5910

you're doing fine

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