## Standard Form to Vertex Form of a Quadratic Equation

Changing a quadratic equation from standard form to vertex form is not too difficult, especially if you understand what a perfect square is. If not, don’t worry, we’ll cover that, too.

We will be using a method called, *Completing the Square*, which takes a quadratic equation and changes it into a perfect square, which as you’ll see, is very easily factorable.

Read through the notes below. Then watch the video and try the practice problems. Be patient with yourself, be a good student taking notes and practicing. You’ll thank yourself later!

This is our first example of an application of a procedure called, Completing the Square. We will refer to Completing the Square as CTS. What CTS does is take a quadratic equation that is NOT a perfect square, and change it into a square, without changing its value.

Idea #1: Squares are Easily Factored

A quadratic that is a square is easily factored. For example, *x*^{2} – 14*x* + 49 is a square, it is (*x* – 7)^{2}.

Idea #2: Pattern in Squares

A quadratic equation, when the leading coefficient is one, that is a perfect square has a pattern. It will always be the case that half of *b* – squared, will equal *c*. In our previous example, half of – 14 is – 7. (-7)^{2} = 49.

Idea #3: Maintaining Value

If you have an equation like *y* = 3*x* – 8, you can add or subtract values from the equation without changing the value of the equation. For example, you could add 2 to each side of the equation.

There is NOT a good reason to add to both sides, but doing so does not change the value of the original equation.

You can also add, and subtract, the same value from one side of the equal sign. For example, we can add and subtract two from the right side.

*y* = 3*x* – 8 + 2 – 2

Pulling It Together

Suppose we had the equation below.

*y* = *x*^{2} + 6*x* – 1

This is NOT factorable, and certainly not a perfect square. For this to be a perfect square, what would the *c* term have to equal? Remember, if it is a perfect square (quadratic) then half of *b* – squared equals *c*.

If *c* = 9, this would be a perfect square. What we can do is add 9 and subtract 9 from the same side of the equation, like seen below.

Here, we have a perfect square, *x*^{2} + 6*x* + 9, and some extra.

We can factor the perfect square, and combine the *extras*, to arrive at the following.

*y* = (*x* + 3)^{2} – 10

This is the vertex form of our original equation, *y* = *x*^{2} + 6*x* – 1. The vertex is (-3, -10).

Summary: To change a quadratic to vertex form, we change it to a perfect square, with a little extra. We take half of *b*, square it, and then add it and subtract it from the same side of the equation. We factor the perfect square and combine extra.

Let’s formalize the process, and talk about two short-cuts.

The example we just completed is pretty straight forward. Things are trickier when *b* is an odd number because when you take half of it, you end up with a fraction. Let’s see an example, and verify that it works. This will help you understand why you do NOT want to square half of b when it is added. That is the factor you’re looking for to factor the perfect square.

First step is to rewrite the equation, leaving space and a + sign after the *bx* term.

Now we take half of b, square it, and add it and subtract it. Do not square the added term, but do square the subtracted term. This helps us carry out our next step more efficiently.

Now, let’s factor the square and simplify the “extra.”

Note: The fraction in the extra: Make this easy. You need a common denominator of four. Well, 16/4 is 4, and 16 – 25 = -9.

Note: If you squared the part you added you’d get the following. To factor this you’d need to find two numbers that multiply to 25/4 and add to make -5. Check if (-5/2) satisfies this. It does. That is hard to figure out though, which is why you don’t square it! Another way to see this is to square *x* – 5/2, to see if you get *x*^{2} – 5*x* + 25/4.