SOLUTION: Hello! I am trying to solve a problem and not sure if its correct. The equation is y= {{{5x^2+10x+7}}}. I must write the quadratic equation into vertex form and find the vertex.

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> SOLUTION: Hello! I am trying to solve a problem and not sure if its correct. The equation is y= {{{5x^2+10x+7}}}. I must write the quadratic equation into vertex form and find the vertex.       Log On

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Question 355007: Hello! I am trying to solve a problem and not sure if its correct. The equation is y= 5x%5E2%2B10x%2B7. I must write the quadratic equation into vertex form and find the vertex.
What I did was factor out the 5 from the equation. So the result would be y=5%28x%5E2%2B2x%29%2B7. To complete the perfect square trinomial, I made the equation y=5%28x%5E2%2B2x%2B1%29%2B7 or 5%28x%2B1%29%5E2%2B7 adding 1 to both sides. Moving the constants to one side the equation came out to be y= 5%28x%2B1%29%5E2%2B6 leaving the vertex to be (-1,6).
Is this correct?
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(28595) About Me  (Show Source):
You can put this solution on YOUR website!
First, we need to complete the square for the expression 5x%5E2%2B10x%2B7




5x%5E2%2B10x%2B7 Start with the given expression.


5%28x%5E2%2B2x%2B7%2F5%29 Factor out the x%5E2 coefficient 5. This step is very important: the x%5E2 coefficient must be equal to 1.


Take half of the x coefficient 2 to get 1. In other words, %281%2F2%29%282%29=1.


Now square 1 to get 1. In other words, %281%29%5E2=%281%29%281%29=1


5%28x%5E2%2B2x%2Bhighlight%281-1%29%2B7%2F5%29 Now add and subtract 1 inside the parenthesis. Make sure to place this after the "x" term. Notice how 1-1=0. So the expression is not changed.


5%28%28x%5E2%2B2x%2B1%29-1%2B7%2F5%29 Group the first three terms.


5%28%28x%2B1%29%5E2-1%2B7%2F5%29 Factor x%5E2%2B2x%2B1 to get %28x%2B1%29%5E2.


5%28%28x%2B1%29%5E2%2B2%2F5%29 Combine like terms.


5%28x%2B1%29%5E2%2B5%282%2F5%29 Distribute.


5%28x%2B1%29%5E2%2B2 Multiply.


So after completing the square, 5x%5E2%2B10x%2B7 transforms to 5%28x%2B1%29%5E2%2B2. So 5x%5E2%2B10x%2B7=5%28x%2B1%29%5E2%2B2.


So y=5x%5E2%2B10x%2B7 is equivalent to y=5%28x%2B1%29%5E2%2B2.


So the equation y=5%28x%2B1%29%5E2%2B2 is now in vertex form y=a%28x-h%29%5E2%2Bk where a=5, h=-1, and k=2


Remember, the vertex of y=a%28x-h%29%5E2%2Bk is (h,k).


So the vertex of y=5%28x%2B1%29%5E2%2B2 is (-1,2).


A graph will confirm this



Graph of y=5x%5E2%2B10x%2B7 with vertex (-1,2)


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim

Answer by stanbon(57361) About Me  (Show Source):
You can put this solution on YOUR website!
y= 5x%5E2%2B10x%2B7. I must write the quadratic equation into vertex form and find the vertex.
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5x^2+10x+7 = y
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5x^2+10x = y - 7
5(x^2+2x+1) = y-7+5*1
--
5(x+1)^2 = y-2
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Vertex: (-1,2)
----
graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C5x%5E2%2B10x%2B7%29}
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Cheers,
Stan H.