Section 2.4: Modeling Quadratic Data
Photo Credit: Shedinagarden — Retrieved from Flickr.com
Recognizing Characteristics of Parabolas
The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. These features are illustrated in Figure 2.4.1 .
Figure 2.4.1: Quadratic descriptive graph
The y-intercept is the point at which the parabola crosses the $y-axis$, $x=0$. The x-interept(s) are the point(s) at which the parabola crosses the $x-axis$, $y=0$. If they exist, the x-intercept(s) represent the zeros, or roots , of the quadratic function. Not all quadratics will cross the $x$-axis. When the graph does not cross the $x$-axis, we call the zeros or roots imaginary.
Example 1
Using the graph below, determine the vertex, axis of symmetry, zeros, and y-intercept of the parabola.
Figure 2.4.2: Quadratic Graph of $y=(x-2)^2-1$
Solution
To identify the x-axis of symmetry and the vertex, look at the graph and identify the "turning point", or place where it changes direction. This is your vertex. In this case, it is the point $(2,-1)$. From this, we can also identify the axis of symmetry, which is $x=2$.
Next, identify the intercepts. The points where it crosses the x-axis are $(1,0)$ and $(3,0)$. Lastly, identify where it crosses the y-axis, which would be the point $(0,3)$. All of these can be seen in the figure 2.4.3 below.
Figure 2.4.3: Quadratic Graph of $y=(x-2)^2-1$ with Characteristics.
Now that you understand the basic characteristics of a quadratic graph, we can look at how we can use quadratics to model data.
Quadratic Models
In the previous lesson, we learned about linear regression and fitting data to a linear model. However, not all data is linear. When it comes to real-world applications, a quadratic model may be more appropriate.
One of America's favorite past times is the game of basketball. On any nice spring day, you could drive through a neighborhood and see people of all ages shooting some hoops. Have you ever thought about the path that ball takes once it leaves a player's hand? Allow us to look at this photo below taken by Dan Meyer.
Image 2.4.1: Picture of path taken by a basketball
If you where to fit this image onto a set of axes, would this be a linear model or a quadratic model? It really doesn't make a straight line as the ball can be seen starting to turn around. This path could very well be quadratic, but how can we tell? First, start by placing a grid over top of this image.
Image 2.4.1: Picture of path taken by a basketball
Now create a table for the points "C" through "E"
| x | 0 | 4 | 6 | 7 | 12 |
|---|---|---|---|---|---|
| y | 10 | 14 | 17 | 18 | 19 |
Table: Data for points C through E
To create a model, we will now use some type of software such as Google sheets or Microsoft Excel. There are plenty of software's, but these are the ones we primarily use in this course. If you are looking for tutorials on how this can be done in Google sheets or Microsoft excel, please see the Appendix.
This also could be done the long way, which would be using quadratic regression and some graduate level statistics skills. For the most part, no one does this the long way. After putting this raw data into Excel, you'd find out that the estimated equation that would best model this data would be $\hat{y}=-0.0631x^2+1.5399x+9.7595$ with a coefficient of variation of $r^2=0.9758$, which would produce a correlation coefficient $r=\sqrt{r^2}=\sqrt{0.9758}=0.9878$ rounded to four decimal places(ten thousandths). Recall in section 1.2 that you learned that the closer $|{r}|$ is to $1$, the stronger the relationship. This correlation coefficient $r$ suggests that there is a rather strong quadratic relationship to our data.
Where would a person use this information and why is it important? A person would use this information if they were trying to make predictions or estimations regarding the data they are interested in. With this particular set of data, our question is, "Will the ball go in the hoop?" To answer this, we will need to go back to graph and find an approximate location for the hoop. Based on our grid, the hoop is approximately at $(22,14)$. Let's check to see if this ball potentially will go in the hoop.
To do this, we need to plug our values into the equation and see if it works out.
\begin{aligned} 14&\stackrel{?}{=}-0.0631(22)^2+1.5399(22)+9.7595 \\ 14&\stackrel{?}{=}-30.5404+33.8778+9.7595\\ 14&\stackrel{?}{=}13.0969\\ \end{aligned}Looking at our solution, $13$ is just slightly below the basket. Thus, it is possible that this ball could go in the hoop at the x-position $22$, the center of the hoop from our graph. This is just an example of one way we can use equations or models to make predictions. Let's look at another example of how models can be applied in the business world.
Example 2
After a rough quarter, a company hires an economic consultant to look at their numbers. The consultant was able to find a model for their data. The consultant found that the profit made $P(x)$ in thousands of dollars from selling $x$ toys (in hundreds) could be approximated by $P(x)=-0.44x^2+4.4x-4$. Using the graph provided, how many toys must the company sell to make a maximum profit? What is their maximum profit, and how many toys must they sell to break-even?
Figure 2.4.4: Company Profits
Solution
First, let's break down this question into its pieces and see what it is asking us to find. It is looking for the number of toys that need to be sold to maximize profits. We know that the maximum or minimum depending on the equation occurs at the vertex of the parabola or quadratic. In this case, our vertex would be the red point below.
Figure 2.4.5: Maximum Company Profits
This would be the point $(5,7)$. Recall that the $x-axis$ represents the number of toys sold in hundreds, and the $y-axis$ represents the profit the company can expect from selling $x$ toys in thousands of dollars. The first part of this question just wanted to know how many toys the company must sell to maximize their profit. Since the $x$ value of our coordinate is $5$ and it is in hundreds, that means the company would need to sell $500$ toys to maximize their profit.The second part of this question asked what is the company's maximum profit from selling those $500$ toys? This would be the $y$ value of our coordinate for the vertex, which is $7$. Remember the $y-axis$ is measured in thousands of dollars, thus the answer would be \$ $7,000$.
The last part of this question asks when will the company break-even. The break-even point is when the profit is zero ($y=0$). This occurs when the graph crosses the $x-axis$ or the $x$ intercepts. Looking back at our graph, the blue points below designate where the graph crosses the $x-axis$.
Figure 2.4.5:Company Breakeven points
These two points are $(1,0)$ and $(9,0)$. This tells us that the company needs to sell either $100$ toys or $900$ toys to break-even, or have a profit of \$$0$. Just as a side note, profit occurs when the revenue from selling and object is higher than the cost of making the object. Thus the break-even point is when it costs the same amount to make $x$ objects as it does to sell $x$ objects.