Graphing and Modeling Quadratic Functions

Objectives

After completing this lesson, you will be able to:

Analyze quadratic functions graphs.
Write quadratic functions in standard form and use the results to graph the function.
Find maximum and minimum values of quadratic functions and use them in real-life applications.

This section assumes that you've studied quadratic functions and their roots previously. Please click here if you wish to review these topics. Quadratic functions can be used to model data and be analyzed in a great variety of real-life applications. In this lesson, we will study the quadratic function graphs and how to use them to model diverse problems. We saw in the section on transformations (click here to review) that the standard form shown below allows us to visualize the graph of a quadratic function by starting with f(x) = x² and performing various transformations to arrive at the graph of a quadratic function in standard form.

Standard Form of a Quadratic Function.

The standard form of a quadratic function is:

$f (x) = a {(x - h)}^{2} + k$

Converting an expressed function in general form to its standar form

Any quadratic function $f (x) = a x^{2} + b x + c$ can be expressed in standard form:

$f (x) = a {(x - h)}^{2} + k$

The way to achieve this is by using the method of completing the squre.

Example 1:

Write the function $f (x) = 2 x^{2} + 8 x + 7$ in standard form.

Solution:

Step 1: Factorize the coefficient of $x^{2}$ . $\begin{matrix} f (x) = 2 x^{2} + 8 x + 7 \\ f (x) = 2 (x^{2} + 4 x) + 7 \end{matrix}$
Step 2: In order for the expression with the parentheses to become a perfect square, we add: ${(\frac{4}{2})}^{2} = 4$ . Then, we should subtract the same value so we do not alter the expression. $\begin{matrix} f (x) = 2 (x^{2} + 4 x) + 7 \\ f (x) = 2 (x^{2} + 4 x + 4 - 4) + 7 \end{matrix}$
Step 3: Regroup the terms: $\begin{matrix} f (x) = 2 (x^{2} + 4 x + 4) + 2 (- 4) + 7 \end{matrix}$
Step 4: Rewrite the expression in parenthesis, which is now a perfect square and simplify it. $\begin{matrix} f (x) = 2 {(x + 2)}^{2} - 8 + 7 \\ f (x) = 2 {(x + 2)}^{2} - 1 \end{matrix}$

Example 2:

Write function $f (x) = - x^{2} + 2 x + 5$ in standard form.

Solution:

Step 1: Factorize the coefficient $x^{2}$ . $\begin{matrix} f (x) = - x^{2} + 2 x + 5 \\ f (x) = - (x^{2} - 2 x) + 5 \end{matrix}$
Step 2: In order for the expression within the parenthesis to become a perfect square, we add: ${(\frac{2}{2})}^{2} = 1$ . Then, we should also subtract the same value so we do not alter the expression. $\begin{matrix} f (x) = - (x^{2} - 2 x) + 5 \\ f (x) = - (x^{2} - 2 x + 1 - 1) + 5 \end{matrix}$
Step 3: Regroup the terms: $\begin{matrix} f (x) = - (x^{2} - 2 x + 1) - (- 1) + 5 \end{matrix}$
Step 4: Rewrite the expression in parenthesis which is now a perfect square and simplify it. $\begin{matrix} f (x) = - {(x + 1)}^{2} + 1 + 5 \\ f (x) = - {(x + 1)}^{2} + 6 \end{matrix}$

Click on the following link to practice the standard form of a quadratic function:

Graph of quadratic functions.

Graph $f (x) = x^{2}$ .

The graph of function: $f (x) = x^{2}$ is the following:

x	$f (x) = x^{2}$
-4	16
-3	9
-2	4
-1	1
0	0
1	1
2	4
3	9
4	16

x al cuadrado

The graph of a quadratic function receives the name, parable. These always have an extreme value called vertex of the parable. This value is very important practical applications when we want to find the maximum or minimum value of a quadratic function, as we observe later on, when we develop the application problems.

Graph $f (x) = a {(x - h)}^{2} + k$

The standard form is very convenient in order to draw the parable, applying transformations to graph $f (x) = x^{2}$ .

Value |a| produces a stretching or vertical shinkage.
If a<0, the graphic is reflected on the x-axis.
Value h represents a horizontal shift of h units.
Value k represents a vertical shift of k units.

The transformation of functions is explained in detail in the tutorial Transformation of Functions.

Example 1:

Graph function $f (x) = 2 {(x - 1)}^{2} - 5$

Solution:

$x^{2} \Rightarrow 2 \times x^{2}$	Graph of $f (x) = x^{2}$	We multiply by two to obtain: $f (x) = 2 x^{2}$
$2 x^{2} \Rightarrow 2 {(x - 1)}^{2}$	Graph of $f (x) = 2 x^{2}$	We substract one to the entry and we obtain: $f (x) = 2 {(x - 1)}^{2}$
$2 {(x - 1)}^{2} \Rightarrow 2 {(x - 1)}^{2} - 5$	Graph of $f (x) = 2 {(x - 1)}^{2}$	Substracting five to the output we have: $f (x) = 2 {(x - 1)}^{2} - 5$
$2 {(x - 1)}^{2} - 5$	Graph of $f (x) = 2 {(x - 1)}^{2} - 5$
$x^{2} \Rightarrow \frac{1}{2} \times x^{2}$
$\frac{1}{2} x^{2} \Rightarrow - \frac{1}{2} x^{2}$
$- \frac{1}{2} x^{2} \Rightarrow - \frac{1}{2} {(x + 2)}^{2}$
$- \frac{1}{2} {(x + 2)}^{2} \Rightarrow - \frac{1}{2} {(x + 2)}^{2} + 8$
$- \frac{1}{2} {(x + 2)}^{2} + 8$


Click on the Large Buttons to Watch the Transformations.

Example 2:

Graph function $f (x) = - \frac{1}{2} {(x + 2)}^{2} + 8$

Solution:


Click on the Large Buttons to Watch the Transformations

Maximum and Minimum Values

By observing the graphs in the previous examples, we see the extreme value of the function occurs right in the vertex, which is easy to distinguish when we have the function written in standard form.

Given a quadratic function f in standard form:

$f (x) = a {(x - h)}^{2} + k$

The maximum or minimum value of f occurs in x=h

If a>0, the minimum value is f(h)=k.
If a<0, the maximum value is f(h)=k.

Example:

Given the following function

f (x) = 5 x^{2} - 30 x + 49

Express f in standard form.
Create an outline of the graph.
Which is the minimum value of f?

Solution:

We express f in standard form.

$\begin{matrix} f (x) = 5 x^{2} - 30 x + 49 \\ f (x) = 5 (x^{2} - 6 x) + 49 \\ f (x) = 5 (x^{2} - 6 x + 9 - 9) + 49 \\ f (x) = 5 (x^{2} - 6 x + 9) + 5 (- 9) + 49 \\ f (x) = 5 {(x - 3)}^{2} - 45 + 49 \\ f (x) = 5 {(x - 3)}^{2} + 4 \end{matrix}$

The formula indicates that the parable has a minimum value, since a=5 is positive. As we observe in the graph, the vertex of the parable is (3,4) and it opens upwards.

The minumum value is f(3)=4.

Modeling

In the following application, we will model the trayectory of a cannonball. This trayectory has the form of a parable. On the right side, we see the quadratic function written in standard form. On the left side, we see three line segments which represent the values that parameters a, h and k can take on, which correspond to this function. Moving the corresponding points, changes the values of these parameters, and therfore, changes the graph.

Test by moving the points and observe the changes produced in the graph and in the equation.

The objective of this exercise is to move the points in order to change parameters a, h y k in such a way that the parable obtained coincides with the trayectory of the cannon, demonstrated in e forma tal que la parábola obtenida coincida con la trayectoria de la cannonball, shown in dotted lines.

Once you have achieved this, you will observe on the right side the equation which corresponds to the cannonball.

This is a Java Applet created with from www.geogebra.org – Java does not appear to be installed in this equipment. We suggest you go to www.java.com

Application Problems

Example 1:

A ball is thrown in a playing field. Its trayectory is given by the function:

$f (x) = - \frac{1}{12} x^{2} + 2 x + 4$

Which is the maximum height the ball reaches?
At what horizontal distance from the launch point was the maximum height reached?
What is the maximum value of f?

Solution:

We express f in standard form.

$\begin{matrix} f (x) = - \frac{1}{12} x^{2} + 2 x + 4 \\ f (x) = - \frac{1}{12} (x^{2} - 24 x) + 4 \\ f (x) = - \frac{1}{12} (x^{2} - 24 x + 144 - 144) + 4 \\ f (x) = - \frac{1}{12} (x^{2} - 24 x + 144) - \frac{1}{12} (- 144) + 4 \\ f (x) = - \frac{1}{12} {(x - 12)}^{2} + 12 + 4 \\ f (x) = - \frac{1}{12} {(x - 12)}^{2} + 16 \end{matrix}$

Since the function represents the height the ball travels, its maximum height is k=16.

The maximum height is reached when x =h=12.

It is also easy to find these values by observing the graph.

Example 2:

Find two positive real numbers a and b whose produce is the maximum, and also satisfy that a + 2b = 24

Solution:

We want to model function p so it expresses a product of a y b.

p = a × b

We have two variables. We can use the second condition to write the variable a in terms of b.

a + 2b = 24 where a = 24-2b

We rewrite our function in terms of b.

$p (b) = (24 - 2 b) \times b = - 2 b^{2} + 24 b$

We express p in standard form.

$\begin{matrix} p (b) = - 2 b^{2} + 24 b \\ p (b) = - 2 (b^{2} - 12 b) \\ p (b) = - 2 (b^{2} - 12 b + 36 - 36) \\ p (b) = - 2 (b^{2} - 12 b + 36) - 2 (- 36) \\ p (b) = - 2 {(b - 6)}^{2} + 72 \\ p (b) = - 2 {(b - 6)}^{2} + 72 \end{matrix}$

By observing the standard form of the function, we see the maximum value is reached when b =6.

Now, we calculate the vaue of the second variabe using the expression a = 24-2b where a=12.

Click on the following link to practice application problems.

Summary

Now that you have completed this lesson, you should be able to:

Analyze graphs of quadratic functions.
Write quadratic functions in standard form and use the results to graph the function.
Find maximum and minimum value of quadratic functions and use them in real-life applications.

Graphing and Modeling Quadratic Functions

Converting an expressed function in general form to its standar form

Graph f x = x 2 .

Graph $f (x) = x^{2}$ .