Quadratic equations may take various forms. You have already seen the standard form: 

$f(x)=a{ x }^{ 2 }+bx+c$

Another common form is called vertex form, because when a quadratic is written in this form, it is very easy to tell where its vertex is located. The vertex form is given by:

 $f(x)=a(x-h)^2+k$

The vertex is $(h,k).$ Note that if the form were $f(x)=a(x+h)^2+k$, the vertex would be $(-h,k).$ The coefficient $a$ as before controls whether the parabola opens upward or downward, as well as the speed of increase or decrease of the parabola. 

Converting From Vertex Form to Standard Form

If you want to convert a quadratic in vertex form to one in standard form, simply multiply out the square and combine like terms. For example, the quadratic

$y=(x-2)^2+1$ 

Can be rewritten as follows:

$\begin{aligned} y&=(x-2)(x-2)+1 \\ &=x^2-2x-2x+4+1 \\ &=x^2-4x+5 \end{aligned}$

Converting From Standard Form to Vertex Form 

It is more difficult to convert from standard form to vertex form. The process is called "completing the square."  

Conversion When $a=1$

Consider the following example: suppose you want to write $y=x^2+4x+6$ in vertex form. Note that the coefficient on $x^2$ (the one we call $a$) is $1$. When this is the case, we look at the coefficient on $x$ (the one we call $b$) and take half of it. Then we square that number. Thus for this example, we divide $4$ by $2$ to obtain $2$ and then square it to obtain $4$. 

We then both add and subtract this number as follows: 

$y=(x^2+4x+4)+6-4$

Note that we both added and subtracted 4, so we didn't actually change our function. Now the expression in the parentheses is a square; we can write $y=(x+2)^2+2.$ Our equation is now in vertex form and we can see that the vertex is $(-2,2).$ 

Conversion When $a \neq 1$

It is slightly more complicated to convert standard form to vertex form when the coefficient $a$ is not equal to $1$. We can still use the technique, but must be careful to first factor out the $a$ as in the following example: 

Consider $y=2x^2+12x+5. $ We factor out the coefficient $2$ from the first two terms, writing this as:

$y=2(x^2+6x) + 5$ 

We then complete the square within the parentheses. Note that half of $6$ is $3$ and $3^2=9$. So we add and subtract $9$ within the parentheses, obtaining:

$y=2(x^2+6x+9-9)+5$ 

We can then finish the calculation as follows:

$\begin{aligned} y&=2((x+3)^2-9)+5 \\ &=2(x+3)^2-18+5 \\ &=(x+3)^2-13 \end{aligned}$ 

So the vertex of this parabola is $(-3,-13).$