anonymous
  • anonymous
Rewrite the equation y = 3x^2 + 6x + 2 in the alternate form and state the vertex point.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
do you see how y = 3x^2 + 6x + 2 is in the form y = ax^2 + bx + c ?
anonymous
  • anonymous
a = 3 b = 6 c = 2?
jim_thompson5910
  • jim_thompson5910
yep

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jim_thompson5910
  • jim_thompson5910
now plug the values of a,b into h = -b/(2a) what is the value of h?
anonymous
  • anonymous
h = -1?
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • 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
  • anonymous
Is it -1?
jim_thompson5910
  • jim_thompson5910
correct
jim_thompson5910
  • jim_thompson5910
so k = -1
jim_thompson5910
  • jim_thompson5910
we have a = 3, h = -1 and k = -1 plug those into y = a(x-h)^2 + k
anonymous
  • anonymous
Does x = 6?
jim_thompson5910
  • jim_thompson5910
where are you getting x = 6 ?
anonymous
  • anonymous
Oh thought x = 6 from the problem. Dont know what x is.
jim_thompson5910
  • jim_thompson5910
x is just a variable to be left alone really
jim_thompson5910
  • jim_thompson5910
what did you get when you plugged those values into y = a(x-h)^2 + k
anonymous
  • anonymous
Is it 11?
jim_thompson5910
  • jim_thompson5910
don't replace x though
anonymous
  • anonymous
Oh keep x as just x?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
So is this right? y - 3(x - (-1)^2 + - 1 . I think I set it up wrong.
anonymous
  • anonymous
I meant y =
jim_thompson5910
  • jim_thompson5910
if you simplified that, you'd get what?
anonymous
  • anonymous
Does the 3 go with the things inside the ( )
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
|dw:1442024648925:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1442024665299:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1442024695181:dw|
jim_thompson5910
  • 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
  • anonymous
Yeah does the 3 multiply with that?
jim_thompson5910
  • 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
  • jim_thompson5910
what does `x-(-1)` simplify to?
anonymous
  • anonymous
3x - (-3)^2 - (-1) ?
jim_thompson5910
  • jim_thompson5910
if you subtract a negative, what are you really doing?
anonymous
  • anonymous
Making it positive?
jim_thompson5910
  • jim_thompson5910
yeah adding so `x-(-1)` is the same as `x+1`
jim_thompson5910
  • jim_thompson5910
making \[\Large y=3(x-(-1))^2+(-1)\] turn into \[\Large y=3(x+1)^2-1\]
anonymous
  • anonymous
Oh okay I see.
jim_thompson5910
  • jim_thompson5910
What would be the vertex here?
anonymous
  • anonymous
I am not really sure.
anonymous
  • anonymous
Do you put everything together?
jim_thompson5910
  • jim_thompson5910
well the general vertex form equation \[\Large y = a(x-h)^2 + k\] has the vertex `(h,k)`
anonymous
  • anonymous
(1, -1) ?
jim_thompson5910
  • jim_thompson5910
well recall that h = -1 and k = -1
anonymous
  • anonymous
Oh thats not the vertex?
jim_thompson5910
  • jim_thompson5910
no but you're close
anonymous
  • anonymous
I do not get how to find it.
jim_thompson5910
  • jim_thompson5910
it's (-1,-1) you start with (h,k) and replace h and k with the values we got earlier
jim_thompson5910
  • jim_thompson5910
h = -b/(2a) = -1 after plugging in a and b k = -1 after plugging x = -1 into the original equation
anonymous
  • anonymous
Oh I see where it came from.
anonymous
  • anonymous
Is the vertex (-1, -1) or is that used to find the vertex?
jim_thompson5910
  • 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
  • anonymous
Oh okay and then y = 3(x + 1)^2 - 1 is the equation rewritten?
anonymous
  • anonymous
in the alternate form*
jim_thompson5910
  • jim_thompson5910
that's vertex form, yes
anonymous
  • anonymous
Oh okay thanks for your help. I am very slow in math as anyone can tell.
jim_thompson5910
  • jim_thompson5910
you're doing fine

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