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.

Question 355007: Hello! I am trying to solve a problem and not sure if its correct. The equation is y= . 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=. To complete the perfect square trinomial, I made the equation y= or adding 1 to both sides. Moving the constants to one side the equation came out to be y= leaving the vertex to be (-1,6).
Is this correct?
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
First, we need to complete the square for the expression

Start with the given expression.

Factor out the coefficient . This step is very important: the coefficient must be equal to 1.

Take half of the coefficient to get . In other words, .

Now square to get . In other words,

Now add and subtract inside the parenthesis. Make sure to place this after the "x" term. Notice how . So the expression is not changed.

Group the first three terms.

Factor to get .

Combine like terms.

Distribute.

Multiply.

So after completing the square, transforms to . So .

So is equivalent to .

So the equation is now in vertex form where , , and

Remember, the vertex of is (h,k).

So the vertex of is (-1,2).

A graph will confirm this

Graph of 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(75887) (Show Source): You can put this solution on YOUR website!
y= . I must write the quadratic equation into vertex form and find the vertex.
-----
5x^2+10x+7 = y
---
5x^2+10x = y - 7
5(x^2+2x+1) = y-7+5*1
--
5(x+1)^2 = y-2
---
Vertex: (-1,2)
----
}
---
Cheers,
Stan H.
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