Completing the Square to find Vertex Form
Going from standard form, f ( x )  ax 2  bx  c to vertex form, f ( x )  a ( x  h) 2  k is done by
completing the square.
First we must know what a squared binomial is. Bi means 2. Nomial is number. Therefore, binomial is
( x  h) . A squared binomial would be ( x  h) 2 , which we see in the vertex form.
If we FOIL the squared binomial we would get a quadratic in standard form.
( x  h) 2
( x  h)( x  h)
x 2  hx  hx  h 2
x 2  2hx  h 2
We can use this pattern to square any binomial.
( x  3) 2  x 2  6 x  9
( x  2) 2  x 2  4 x  4
So our a term is our x value squared, our b value is our h term doubled times x , and our c term is
our h term squared.
Because of this we can undo or unsquared a perfectly square quadratic.
because our h term is half our b term and our c term is h 2 we can see if it’s a perfect square.
f ( x)  x 2  2 x  1
2
 1,12  1
2
perfect.
( x  1) 2  x 2  2 x  1
To transfer from a non-perfect square quadratic, we have to make a perfect square.
f ( x)  x 2  4 x  2
quadratic in standard form
f ( x)  2  x  4 x
move the constant first
f ( x)  2  4  x  4 x  4
make a perfect square quad. By taking half the b term,
2
2
squaring it and adding it to both sides.
f ( x)  2  ( x  2)
2
simplify the side with the constants and undo the perfect
square to make it a binomial.
f ( x)  ( x  2)2  2
Now move the constant back, and your quadratic is now in
vertex form.
You follow the same steps if your a term is not one but you must add a couple steps.
f ( x)  3 x 2  6 x  2
Quadratic in standard form.
f ( x)  2  3 x 2  6 x
Move the constant.
f ( x)  2  3( x 2  2 x)
Now we have to divide by the a term. But leave the a term on
the same side.
f ( x)  2  3(1)  3( x  2 x  1)
2
Now within the () create the perfect square quadratic by taking
half the b term, squaring it and adding to both sides. Be careful
when adding to the opposite side you must multiply by the a
term factor you removed earlier.
f ( x)  5  3( x  1)
2
Simplify the one side and create the squared binomial on the
other, making sure to leave the factored out a term in front.
f ( x)  3( x  1)  5
2
Bring back the constant, and now your quadratic is in vertex
form and has a vertical stretch.
2