We explore the vertex form of a quadratic.

The Vertex Form of a Quadratic

We have learned the standard form of a quadratic function’s formula, which is . But quadratic functions also have different forms, similar to linear functions. Here, we will learn another form for quadratic functions called the vertex form.

Using graphing technology, consider the graphs of and on the same axes.

We see only one parabola because these are two different forms of the same function. Indeed, if we convert into standard form:

it is clear that and are the same function. We see that the graphs of the two parabolas are completely identical.

The formula given for is said to be in vertex form because it allows us to read the vertex without doing any calculations. The vertex of the parabola is . We can see those numbers in . The -value is the solution to , and the -value is the constant added at the end.

Notice that the -coordinate of the vertex has the opposite sign as the value in the function formula. On the other hand, the -coordinate of the vertex has the same sign as the value in the function formula. Let’s look at an example to understand why. We will evaluate .

The -value is the solution to , which is positive . When we substitute for we get the value . Note that these coordinates are the coordinates of the vertex at . Now we will see how to rewrite quadratics given to us in standard form in vertex form.

Converting to Vertex Form by Completing the Square

In order to convert a quadratic from standard form to vertex form, we will use a technique called completing the square. Consider a generic , with , and real numbers, and assume that . We assume this because if were zero, we would have a linear function instead of a quadratic one, which is not the focus here. The quadratic formula used to find the values of for which or, in other words, the possible -intercepts, is usually a source of grief for people studying quadratic functions for the first time. Let’s try to remedy this by understanding where such a formula comes from. The main strategy here is a little algebraic device called “completing the squares”, which is also useful for finding the coordinates of the vertex of the parabola given by the graph of .

Very briefly, the procedure consists in noting that and paying close attention to the term. Comparing this with the linear term you were given will tell you what should be. Add and subtract whatever you need to in the quadratic function you were given, to produce (or, more generally, the multiple ), and whatever constant term is left outside the factored will be the desired . We’ll see several examples below.

Very generally, consider , with , and real numbers, with . Let’s repeat everything we have done in the previous example, with , and instead of concrete numbers. Here are the steps we can follow:

  • First, look only at .
  • Compare with . The we need here is . Note that .
  • Because of the we had to factor out in the beginning, let’s add and subtract from the original expression.
  • Compute
  • Hence, the vertex of the parabola described by is given by

And with those computations in place, we are in fact very close to understanding where the quadratic formula came from. Solving is, by the above, the same as solving Reorganize as and divide by to get Assuming that , we may take square roots on both sides: Eliminating the absolute values, we have Now solve for , by putting everything on the right side under a common denominator: Mystery solved. For each choice of sign , we get one -intercept. Now, we also observe that the average of such solutions does give the -coordinate of the vertex, as you might expect:

The -coordinate of the vertex will be, naturally, . This can also be used as a shortcut to write quadratic functions in vertex-form.