Question 1042680: Hi, I'm having a really difficult time understanding how to convert a Quadratic from Standard form to Vertex form. I've watched videos and have read multiple websites explaining it but either the videos are too fast, or the lesson is too confusing. I was wondering if there would be any way you could explain the steps in converting standard form to vertex form as simple as possible? Thank you so so much!!
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Standard is ax^2+bx+c=0
Start with x^2-2x-6=0
The vertex x value is -b/2a
Here, that would be -(-2)/2(1)=2/2=1
So the x-value is 1
Plug that into the equation and the y value is 1^2-(2)(1)-6=1-2-6=-7
The vertex is (1,-7)
The vertex form is y-=a(x-h)+k. Change the sign of the x coordinate and keep the sign of the y.
y=(x-1)^2-7
You can always check by foiling out the square term.
It is x^2-2x+1-7=x^2-2x-6
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3x^2-6x-5=0
divide the first two terms by 3 and put the 3 out front.
3(x^2-2x)-5=0
complete the square by taking half the x term and squaring it. Whatever it is, subtract it from the constant to keep the equation balanced. We have added a 1 inside the parentheses and the whole inside term is multiplied by 3. We have to subtract 3 from the equation, too.
3(x^2-2x+1)-5-3=3(x^2-2x+1)-8
Now write it as the square.
3(x-1)^2-8
The vertex is at (1,-8), change the sign of the number after the x and keep the sign of the constant at the end.
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! explain the steps in converting standard form to vertex form
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y = ax^2 + bx + c
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Factor out "a" to get::
y = a(x^2 + (b/a)x) + ? + c
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Complete the square and maintain the equality of the sides::
y = a(x^2 + (b/a)x + (b/2a)^2) + c - a(b/2a)^2
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Factor to get::
y = a(x+(b/2a))^2 + c - (b^2/4a)
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Simply::
y = a(x+(b/2a))^2 + [4ac-b^2]/(4a)
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Rearrange::
y = a(x+(b/2a))^2 - [b^2-4ac]/(4a)
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Cheers,
Stan H.
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