How do you write this in the quadratic equations in vertex form? y=x^2-4x+6

Question

anonymous · Answer

You need to use the method called "Completing the square". 
Given 
$$y=x^2 +bx +c$$
you can complete the square by 
$$y=x^2 +bx +\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 + c$$
note that by adding and subtracting the (b/2)^2 term I haven't changed the equation at all (like adding 5 and subtracting 5 doesn't change a value) 
But we know that 
$$\left(x+\frac{b}{2}\right)^2= x^2 + \frac{b}{2}x+\frac{b}{2}x +\left(\frac{b}{2}\right)^2 $$
$$\left(x+\frac{b}{2}\right)^2= x^2 + 2\frac{b}{2}x+\left(\frac{b}{2}\right)^2 $$
$$\left(x+\frac{b}{2}\right)^2= x^2 + bx+\left(\frac{b}{2}\right)^2 $$
so 

$$y=x^2 +bx +\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 + c$$
therefore becomes

$$y=\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 + c$$
which is essentially in the vertex form of y=k(x-p)^2+q
where 
k=1, p=-(b/2), and q= c-(b/2)^2

Note that if you start with something like
$$y=ax^2 + 4x+13$$
that you should divide out by a first so you have 
$$\frac{y}{a}= x^2 + \frac{4}{a}x+\frac{13}{a}$$ 
do the process and then at the END multiply through by a. That will give you the 
$$y=k(x-p)^2+q$$ which is the vertex form of a quadratic that gives the vertex at (p,q) 
the axis of symmetry at x=p and whether it is open up (k is positive) or open down (k is negative)

anonymous · Answer

So for example if I had
$$y=x^2 -100x +1$$
I would then write 
$$y=x^2 -100x +\left(\frac{-100}{2}\right)^2-\left(\frac{-100}{2}\right)^2 +1$$
$$y=x^2 -100x +\left(-50\right)^2-\left(50\right)^2 +1$$
and then 
$$y=(x-50)^2 -2500 +1$$

$$y=(x-50)^2 -2499$$
So the vertex would be at (50,-2499), an axis of symmetry at x=50, and it would be open up.