Answer:
y=(ax+(b/2))²+c-(b/2)²
Explanation:
No question was given so I will give an example instead. Lets say that we have the equation y=x²-2x+3 and we want to find the quadratic function in vertex form by completing the square.
y=x²-2x+3 follows the form y=ax²+bx+c. Using the coefficients, we see that a=1, b=-2, and c=3.
To complete the square we want to isolate our x's. We can show this by putting parenthesis around our x terms or by moving the 3 to the other side.
y-3=(x²-2x)
Now, we want to complete the square of the expression x²-2x. We know that squares follow the formula (a+b)²=a²+2ab+b².
Using this equation we can find that a=1 and -2=2ab.
We simplify to -2=2b
b=-1
Let's square this to follow the formula a²+2ab+b², now we get
y-3=(x²-2x+1)
However if we add 1 to one side, then we have to add it to the other, so the real equation is
y-3+1=x²-2x+1
y-2=x²-2x+1
We can now factor x²-2x+1 to (x-1)², so
y-2=(x-1)²
y=(x-1)²+2, which is our final answer in vertex form.
Simply put, in the equation y=ax²+bx+c, to complete the square you would divide b by 2 (b/2) and then square it (b/2)². Then you would add it to both sides of the equation.
y+(b/2)²=ax²+bx+(b/2)²+c
This factors into
y+(b/2)²=(ax+(b/2))²+c
y=(ax+(b/2))²+c-(b/2)²