Instructional Unit 6

Standard Form of a Quadratic Function

by Sandy Cederbaum

The Standard Form of a quadratic function (often referred to as polynomial form) is given by . We can expand any of the vertex form quadratic functions that we worked with in unit 5 and write it in standard form. To do this, we must be able to expand a binomial square (FOIL) and follow the order of operations properly. Here is an example of a function that we worked with in unit 5 expanded and converted into Standard Form:

where a=-2, b=12, and c=-13. These coefficients will have greater meaning to us soon enough.

Expand each of the four remaining vertex form functions from Unit 5 , transform each into standard form, and state the values of the coefficients a, b, and c for each.

In our study of quadratic functions and equations, we will often be given a quadratic function in standard form. We may be asked any of the following questions:

1. Find the y-intercept of the function.

2. Find the value of y (or the dependent variable) when x (the independent variable) is specified.

3. Find the x-intercepts of the function.

4. Find the vertex (or find the coordinates of the maximum or minimum value on the graph) of the function.

5. Find the value of x (or the independent variable) when y (the dependent variable) is specified.

Let us use the function in order to address each of these questions. We will assume that we do not already know the vertex form of this function.

The first two questions above should not be too much trouble to solve given a quadratic function in standard form. However, the latter three questions are a bit more challenging. One tactic we could try in order to find the x-intercepts is to use the graph of the function to try to find the vertex form equation of the quadratic.

Since the vertex appears to be at (3,5), we might start with a function that looks like . Since the coefficient of the squared term is -2, we might try a = -2. In order to see why this makes sense, expand f(x).

If we now compare the two functions and , a=-2, -6a=12, 9a+5=-13. Each of these three equations lead us to the conclusion that a=-2. Now we can use the vertex form function and what we learned in the previous units in order to find the x-intercepts by setting f(x)=0.

As we already have the vertex, we can move on to the fifth question above. For example, find the value(s) of x when f(x)=2. We can get a decent approximation for these values simply by looking at the graph. In order to solve this problem algebraically, we could take a similar tact as we did to find the x-intercepts. We are looking at the equation . So by subtracting 2 from both sides of the equation, we end up with a new quadratic function. The purple function below is the graph for which we have already found the x-intercepts. The blue graph is the new function. The zeros of this function represent the solutions of the equation .

Note: the entire graph has shifted down 2 units. There is much to be discovered here if you continue to play with the value of f(x). For example, what value of f(x) will return only one x value? What values of f(x) will not return a real x value?

We can continue as we did and transform this function into vertex form and find the values of x that make this equation true. This is not a particularly elegant or efficient technique, and sometimes it can be difficult to determine the vertex of the parabola. In the next unit, we will learn a process called completing the square that will enable us to develop a routine to convert any standard form quadratic function to vertex form. This process will lead us to some nice shortcuts to finding the vertex and x-intercepts of quadratic functions.

Move on to Unit 7 - Completing the Square