SOLUTION: Write y=2x^2+12x+14 in vertex form

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Question 112191: Write y=2x^2+12x+14 in vertex form
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
y=2x%5E2%2B12x%2B14 Start with the given quadratic


y-14=2x%5E2%2B12x Subtract 14++ from both sides


y-14=2%2A%28x%5E2%2B6%2Ax%29 Factor out the leading coefficient 2++. This step is important since we want the x%5E2 coefficient to be 1.


Take half of the x coefficient 6 to get 3 (ie 6%2F2=3)

Now square 3 to get 9 (ie %283%29%5E2=9)



y-14=2%2A%28x%5E2%2B6%2Ax%2B9%29 Add this number (9) to the expression inside the parenthesis. Now the expression x%5E2%2B6%2Ax%2B9 is a perfect square trinomial.


y-14%2B%282%29%289%29=2%2A%28x%5E2%2B6%2Ax%2B9%29 Since we added 9 inside the parenthesis, we need to add %282%29%289%29 to the other side (remember, we factored out the leading coefficient 2).


y-14%2B%282%29%289%29=2%2A%28x%2B3%29%5E2 Factor x%5E2%2B6%2Ax%2B9 into %28x%2B3%29%5E2. Now the right side is a perfect square (note: if you need help with factoring, check out this solver)


y%2B4=2%2A%28x%2B3%29%5E2 Multiply 2 and 9 to get 18.



y=2%2A%28x%2B3%29%5E2-4 Subtract 4 from both sides


Now the equation is in vertex form