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Quadratic Function Models

GETTING STARTED Many college-bound high school students enroll in Advanced Placement Program(AP) course to earn college credit while still in high school. The College Board offer 35 course in 19 disciplines ranging from Studio Art: 3D Design to Calculus AB test has increased dramatically since 1969, as shown in Figure 2.1.

圖 2.1:
\includegraphics[width=3in]{F2-1.eps}

How many students will take the test in 2005? Nobody knows; however, using a mathematical model, we can predict what may happen.

In this section, we will discuss the identifying features of quadratic functions. We will then demonstrate how to use algebraic methods and quadratic regression to model data sets whose graph open upward or downward over the domain.

A polynomial is a function that is the sum of terms of the form $ ax^n$, where $ a$ is a real number and $ n$ is a nonnegative integer. For example, each of the following function is a polynomial.

\begin{displaymath}
\begin{aligned}
f(x)&=6x^4-3x^3+5x^2-2x+9 \\
g(x)&=4x^2+2x \\
h(x)&=-1.3x+9.7 \\
j(x)&=x^2-2x+6
\end{aligned}\end{displaymath}

May a constant term like 9 be written in the form $ ax^n$? Yes! Since $ x^0=1$ for nonzero $ x$, $ 9x^0=9$. Consequently, a constant function $ s(x)=9$ is also a polynomial.

The degree of a polynomial is the value of its largest exponent. For example, the degree of the polynomial $ f(x)=6x^4-3x^3+5x^2-2x+9$ is 4, since 4 is the largest exponent. Since the equation of any line may be written as $ y=ax+b$, lines are polynomials of degree 1. We've worked extensively with first-degree polynomials (lines) in the preceding chapter. In this section, we are intersted in polynomials of degree 2. Polynomials of degree 2 are called quadratic functions .

QUADRATIC FUNCTION
A polynomial function of the form $ f(x)=ax^2+bx+c$ with $ a\ne 0$ is called a quadratic function . The graph of a quadratic function is a parabola .

When a parabola opens upward, we say that it is "concave up." When the parabola opens downward, we say that it is "concave down." The steepness of the sides of the parabola and its concavity are controlled by the value of $ a$, the coefficient on the $ x^2$ term in its equation. If $ a>0$, the graph is concave up. If $ a<0$, the graph is concave down. As the magnitude of $ a$ increases, the steepness of the graph increases. (For $ a>0$, the magnitude of $ a$ is $ a$. For $ a<0$, the magnitude of $ a$ is $ -a$.) Consider the graphs in Figures 2.2 and 2.3. These graphs have the same values for $ b$ and $ c$ ($ b=-2$ and $ c=3$) but differing values of $ a$.

In Figure 2.2, since $ a$ is the coefficient on the $ x^2$ term , $ a = 1$. The magnitude of $ a$ is 1. Since $ a>0$, the graph is concave up.

圖 2.2:
\includegraphics[width=2.4in]{F2-2.eps}

In Figure 2.3, since $ a$ is the coefficient on the $ x^2$ term , $ a=2$. The magnitude of $ a$ is 2. Increasing the magnitude of $ a$ increased the graph's steepness.

圖 2.3:
\includegraphics[width=2.4in]{F2-3.eps}

Consider the graphs in Figures 2.4 and 2.5. These graphs have the same values for $ b$ and $ c$ ($ b = 4$ and $ c=0$) but differing values of $ a$.

In Figure 2.4, since $ a$ is the coefficient on the $ x^2$ term , $ a=-1$. The magnitude of $ a$ is 1. Since $ a<0$, the graph is concave down.

圖 2.4:
\includegraphics[width=2.4in]{F2-4.eps}

In Figure 2.5, since $ a$ is the coefficient on the $ x^2$ term , $ a=-2$. The magnitude of $ a$ is 2. Increasing the magnitude of $ a$ from 1 to 2 increased the graph's steepness.

圖 2.5:
\includegraphics[width=2.4in]{F2-5.eps}

The coefficient $ b$ in the equation affects the horizontal and vertical placement of the parabola. The constant term $ c$ in the equation indicates that the point $ (0,c)$ is the $ y$-intercept of the graph.

Recall that the graph of a function is said to be increasing if the value of $ y$ gets bigger as the value of $ x$ increases. Similarly, the graph of a function is said to be decreasing if the value of $ y$ gets smaller as the value of $ x$ increases. The vertex of a parabola is the point on the graph of a quadratic function where the curve changes from decreasing to increasing (or vice versa). The minimum $ y$ value of a concave up parabola occurs at the vertex (see Figure 2.6a). Similarly, the maximum $ y$ value of a concave down parabola occurs at the vertex (see Figure 2.6b).

圖 2.6:
\includegraphics[width=2.4in]{F2-6a.eps}

圖 2.7:
\includegraphics[width=2.4in]{F2-6b.eps}

For parabolas with $ x$-intercepts. Recall that as a results ot the quadratic formulas, we knows that the $ x$ coordinate of the vertex is $ x=\frac{-b}{2a}$. The $ y$ coordinate of the vertex is obtained by evaluating the quadratic function at this $ x$ value. For example, for the quadratic function $ y=x^2-2x+3$, we know that $ a = 1$ and $ b=-2$. The $ x$ coordinate of the vertex is

\begin{displaymath}
\begin{aligned}
x&=\frac{-b}{2a}\\
&=\frac{-(-2)}{2(1)}\\
&=\frac{2}{2}\\
&=1
\end{aligned}\end{displaymath}

Evaluating the function at $ x=1$ yields

\begin{displaymath}
\begin{aligned}
y&=^2-2x+3 \\
&=(1)^2-2(1)+3 \\
&=1-2+3 \\
&=2
\end{aligned}\end{displaymath}

Therefore, the vertex of the function is $ (1,2)$.

Parabolas are symmetrical. That is, if we draw a vertical line through the vertex of the parabola, the portion of the graph on the left of the line is the mirror image of the line. The line is referred to as the axis of symmetry (see Figure 2.7).

圖 2.8:
\includegraphics[width=2.4in]{F2-7.eps}

Since the axis of symmetry is a vertical line passing through the vertex, the equation of the axis of symmetry is $ x=\frac{-b}{2a}$.


\begin{example}
{\bf Describing the Graph of a Quadratic Function from Its Equat...
...intercept, and the vertex of the quadratic function $y=3x^2+6x-1$.
\end{example}

解: We have $ a=3$, $ b=6$, and $ c=-1$. Since $ a>0$, the parabola is concave up. Since $ c=-1$, $ y$-intercepts is $ (0,-1)$. The $ x$ coordinate of the vertex is given by

\begin{displaymath}
\begin{aligned}
x&=\frac{-b}{2a} \\
&=\frac{-6}{2(3)}\\
&=\frac{-6}{6} \\
&=-1
\end{aligned}\end{displaymath}

The $ y$ coordinate of the vertex is obtained by evaluating the function at $ x=-1$.

\begin{displaymath}
\begin{aligned}
y&=3x^2+6x-1\\
&=3(-1)^2+6(-1)-1\\
&=3-6-1\\
&=-4
\end{aligned}\end{displaymath}

The vertex of the parabola is $ (-1,-4)$.


\begin{example}
{\bf Determining the Equation of a Parabola from Its Graph}\\
Determine the equation of the parabola shown in Figure 2.8.
\end{example}

圖 2.9:
\includegraphics[width=2.4in]{F2-8.eps}

解: The $ y$-intercepts is $ (0,20)$, so $ c=20$. The vertex is $ (3,2)$. Since $ x$ coordinate of the vertex is $ x=\frac{-b}{2a}$, we know that

\begin{displaymath}
\begin{aligned}
\frac{-b}{2a}&=3 \\
-b&=6a \\
b&=-6a
\end{aligned}\end{displaymath}

Since $ f(x)=ax^2+bx+c$, we have

\begin{displaymath}
\begin{aligned}
f(x)&=ax^2+(-6a)x+20 \quad {\text {Since}} \quad b=-6a\\
&=ax^2-6ax+20
\end{aligned}\end{displaymath}

The vertex is $ (3,2)$, so $ f(3)=2$. Therefore,

\begin{displaymath}
\begin{aligned}
f(x)&=ax^2-6ax+20\\
2&=a(3)^2-6a(3)+20 \qu...
...3,2)\\
2&=9a-18a+2 \\
-18&=-9a \\
a&=2
\end{aligned}\\
\end{displaymath}

Since $ -b=6a$, $ b=-12$. The equation of the parabola is $ f(x)=2x^2-12x+20$.

We can check our work by substituting in a different point from the graph. The parabola passes through the point $ (4,4)$. Consequently, this point should satify the equation $ f(x)=2x^2-12x+20$.

\begin{displaymath}
\begin{aligned}
f(x)&=2x^2-12x+20\\
4&=2x^2-12x+20 \quad {...
...4)\\
4&=?2(16)-48+20\\
4&=?32-48+20\\
4&=4
\end{aligned}\end{displaymath}

Recognizing the relationship between the graph of a parabola and its corresponding quadratic function provides a quick way to evaluate the accuracy of a quadratic model. Quadratic models may be determined algebracally or by using quadratic regression. We will demonstrate both methods in the next several examples.

Let's return to the AP Calculus exam data introduced at the beginning of the section. At first glance, the Calulus AB exam data don't look at all like a parabola; however, we do observe that the data appear to be increasing at an ever-increasing rate. (It took six years [1980 to 1986] for the number of exams to increase from 20,000 to 40,000, but it took only three years [1986 to 1989] for the number of exams to increase from 40,000 to 60,000.) Plotting the quadratic equation $ f(t)=157.8486t^2-622,378t+613,500,479$ together with the data set, we observr that, in fact, a quadratic equation fits the data very well (Figure 2.9).(Source:The College Board.) This equation is found by performing quadratic regression on the data, a process that will detail in the forthcoming Technology Tip.

圖 2.10:
\includegraphics[width=3in]{F2-9.eps}

Observe that the coefficients of the variables in the quadratic equation are very large. We can come up with an equation with smaller coefficients by aligning the data . We'll let $ t=0$ in 1969,$ t=1$ in 1970, and so on. Doing quadratic regression on the aligned data yields the equation $ f(t)=157.85t^2-770.64t+10,268$ with the coefficient of determination $ r^2=0.9979$. Recall the coefficient of determination is a measure of how well the model fits the data. The closer$ r^2$ is to 1, the better the model fits the data. Although both of the models fit the data, the second model will make computations easier because of the smaller coefficients.

Using the model, we predict how many tests will be administered in 2005.In 2005, $ t=36$.

\begin{displaymath}
\begin{aligned}
f(36)&=157.85(36)^2-770.64(36)+10,268\\
&=187,099
\end{aligned}\end{displaymath}

We estimate that 187,099 AP Calculus AB exams will be administered in 2005.

A quadratic function model for a data set may be generated by using quadratic regression on a graphing calculator or 3 within Microsoft Excel. However, the fact that technology may be used to create a quadratic model does not guarantee that the model will be a good fit for the data.

Although the quadratic model fits the AP Calculus AB exam data very well from 1969 to 2002, we must be cautious in using it to predict future behavior. (Predicting the output value for an input value outside of the interval of the input data is called extrapolation .) For this data set, we may feel reasonably comfortable with an estimate two or three years beyond the last data point; we would doubt the accuracy of the model 100 years beyond 2002. For example, in 2102 $ (t=133)$, the estimated number of exams is 2,702,587. This figure is more than 17 times greater than the maximum number of exams that have ever been administered!

We also need to look at the population of the students who could possibly take the exam. Between 1970 and 1999, the number of twelfth graders (those who typically take the AP Calculus exam) fluctuated from a high of 3,026,000 in 1977 to a low of 2,381,000 in 1990. (Source: Statistical Abstract of the United States, 2001, Table 232,p.149.) If the number of twelfth graders continues to fluctuate between these two values, at some point our model estimate for the exams would exceed the number of people who could conceivably take the exam. This illustrates the necessity to verify that a model makes sense in its real-world context.

It is often possible to model a data set with more than one mathematical model. When selecting a model, we should consider the following:

  1. The graphical fit of the data
  2. The correlation coefficient $ (r)$ or the coefficient of determination $ (r^2)$
  3. The known behavior of the thing being modeled

Recall that the closer the correlation coefficient is to 1 or -1, the better the model fits the data. Similarly, the closer the coefficient of determination is to 1, the best the model fits the data.


\begin{example}{\bf Using Quadratic Regression to Forecast Prescription Drug Sal...
...e United States increased from 1995 to 2000 as shown in Table 2.1.
\end{example}


表 格 2.1:
Years Since Retail Sales
1995 $ (t)$ (billions of dollars)
  $ [S(t)]$
0 68.6
2 89.1
3 103.0
4 121.7
5 140.7

Source: Statistical Abstract of the United States, 2001, Table 1324, p.829.

Model the data using a quadratic function. Then use the model to predict retail prescription drug sales in 1996 and 2001.

解: We observe from the scatter plot the data appear concave up everywhere (Figure 2.10).

圖 2.11:
\includegraphics[width=2.4in]{F2-10.eps}

A quadratic function may fit the data well. We use quadratic regression to find the quadratic model that best fit the data.

Based on data from 1995 to 2000, retail prescription drug sales in the United States may be modeled by

$\displaystyle S(t)=1.411t^2+7.411t+68.55 \quad {\text {billion}} \quad {\text {dollars}}$

where $ t$ is the number of years since 1995.

The coefficient of determination $ (r^2=0.9997)$ is extremely to 1. The graph also appears to "touch" each data point. The model appears to fit the data well.

In 1996, $ t=1$, and in 2001, $ t=6$. Evaluating at each $ t$ values, we get

\begin{displaymath}
\begin{aligned}
S(1)&=1.411(1)^2+7.411(1)+68.55\\
&=77.402\\
S(6)&=1.411(6)^2+7.411(6)+68.55\\
&=163.992
\end{aligned}\end{displaymath}

Since the original data were accurate to only one decimal place, we will round our solutions to one decimal place as well. We estimate that prescription drug sales were $77.4 billion in 1996 and $164.0 billion in 2001.

You may ask, "Is there a way to find a quadratic model without using quadratic regression?" There is. The model may not be the model of best fit, but it may still model the data effectively. In Example 4, we repeat Example 3 using an algebraic method to find a quadratic model.


\begin{example}{\bf Using Algebraic Methods to Model Prescription Drug Sales}\\ ...
...e United States increased from 1995 to 2000 as shown in Table 2.2.
\end{example}


表 格 2.2:
Years Since Retail sales
1995 $ (t)$ (billions of dollars)
  $ [S(t)]$
0 68.6
2 89.1
3 103.0
4 121.7
5 140.7

Source: Statistical Abstract of the United States, 2001, Table 127, p.94.

Model the data using a quadratic function. Then use the model to predict retail prescription drug sales in 1996 and 2001.

解: Given any three data points, we can find a quadratic function that passes through the points, provided that the points define a nonlinear function. We will pick the points $ (0,68.6)$, $ (3,103.0)$, and $ (5,140.7)$ from the table. A quadratic function is of the form $ {S(t)=at^2+bt+c}$. Each of the points must satisfy this equation.

$\displaystyle \begin{aligned}
68.6&=a(0)^2+b(0)+c \quad &{\text {Substitude}} \...
... &{\text {Substitude}} \quad t=5, S(t)=140.7\\
140.7&=25a+5b+c
\end{aligned}$

Since we know $ c=68.6$, we can simplify the last two equations.

\begin{displaymath}
\begin{aligned}
103.0&=9a+3b+68.6\\
34.4&=9a+3b\\
140.7&=25a+5b+68.6\\
72.1&=25a+5b
\end{aligned}\end{displaymath}

We can now find the values of $ a$ and $ b$ by solving the system of equations.

\begin{displaymath}
\begin{aligned}
9a+3b&=34.4\\
25a+5b&=72.1
\end{aligned}\end{displaymath}

$\displaystyle \begin{bmatrix}
9&3&34.4 \\
25&5&72.1
\end{bmatrix}$

$\displaystyle \begin{bmatrix}
9&3&34.4 \\
-30&0&-44.3
\end{bmatrix}$

$\displaystyle 5R_1-3R_2$

$\displaystyle \begin{bmatrix}
0&30&211.1 \\
-30&0&-44.3
\end{bmatrix}$

$\displaystyle 10R_1+3R_2$

\begin{displaymath}
\begin{aligned}
-30a&=-44.3 \qquad 30b&=211.1 \\
a&=1.477 \qquad b&=7.037
\end{aligned}\end{displaymath}

A quadratic model for the data is

$\displaystyle S(t)=1.477t^2+7.037t+68.6$

Graphing this model with the data shows that it fits the data relatively well (Figure 2.11).

圖 2.12:
\includegraphics[width=2.4in]{F2-11.eps}

In 1996, $ t=1$, and in 2001, $ t=-6$. Evaluating the function at each $ t$ value, we get

\begin{displaymath}
\begin{aligned}
S(1)&=1.477(1)^2+7.037(1)+68.6 \\
&\thickapprox 77.1 \quad {\text {billion}}
\end{aligned}\end{displaymath}

\begin{displaymath}
\begin{aligned}
S(6)&=1.477(6)^2+7.037(6)+68.6 \\
&\thickapprox 164.0 \quad {\text {billion}}
\end{aligned}\end{displaymath}

These estimates are close to the estimates from Example 3. In Example 3, we estimated that prescription drug sales were $77.4 billion in 1996 and $164.0 billion in 2001.


\begin{example}{\bf Using Quadratic Regression to Forecast Nursing Home Care Cos...
...me care has risen substantially since 1960, as shown in Table 2.3.
\end{example}


表 格 2.3:
Years Nursing Home Care
Since 1960 (billions of dollars) [$ N(t)$]
$ (t)$  
0 1
10 4
20 18
30 53
40 96

Source: Statistical Abstract of the United States, 2001, Table 119, p.91.

The U.S. Centers for Medicare and Medicaid Services predict that by 2010 the cost of nursing home care will reach $183 billion.

Model the data with a quadratic function and calculate the cost of nursing home care in 2010. Compare your result to the U.S. Centers for Medicare Medicaid Services estimate.

解: Using quadratic regression, we determine that the quadratic model of best fit is

$\displaystyle N(t)=0.07214t^2-0.4957t+1.029 \quad {\text {billion}} \quad {\text {dollars}}$

where $ t$ is the number of years since 1960.

Since the coefficient of determination $ (r^2=0.9987)$ is extremely close to 1, we anticipate that the model fits the data well. Graphing the data and the model together yields Figure 2.12.

圖 2.13:
\includegraphics[width=2.4in]{F2-12.eps}

The graph passes near each data point. The model appears to fit the data fairly well. Zooming in, however, we see that the graph decreases from 1 billion to 0 billion between 1960 and 1963 (Figure 2.13). (We rounded to whole numbers, since the original data were in whole numbers.)

圖 2.14:
\includegraphics[width=2.4in]{F2-13.eps}

In reality, nursing home expenditures have increased every year since 1960. Neverless. despite this limitation in our model, we will use it to predict nursing home expenditures in 2010. In 2010, $ t=50$.

\begin{displaymath}
\begin{aligned}
N(50)&=0.721(50)^2-0.4957(50)+1.0286\\
&=156.60
\end{aligned}\end{displaymath}

We estimate that in 2010, $157 billion will be spent on nursing home care. Our model estimate is substantially less than the $183 billion that the U.S. Centers for Medicare Medicaid Services estimate. Their estimate probably antipated the health care needs of the aging population of baby boomers, while ours did not.

Models projecting health care costs are useful for legislators, insurance companies, and consumers. By considering future costs, people can prepare for the future and avert financial crises.


\begin{example}{\bf Using Algebraic Methods to Find a Quadratic Model for Net Sa...
...atic model for the net sales of the Kellogg Company algebraically.
\end{example}


表 格 2.4:
Years Since 1999 Kellogg Company Net Sales
$ (t)$ (millions of dollars)
  [$ R(t)$]
0 6,984.2
1 6,954.7
2 8,853.3

Source: Kellogg Company 2001 Annual Report, pp.7,27.

解: Since we are given the $ y$-intercept, (0,6984.2), we know that $ c=6984.2$. We have

\begin{displaymath}
\begin{aligned}
R(t)&=at^2+bt+c\\
&=at^2+bt+6984.2\\
R(1...
...984.2\\
8853.3&=4a+2b+6984.2\\
1869.1&=4a+2b
\end{aligned}\end{displaymath}

We must solve the system of equations

\begin{displaymath}
\begin{aligned}
a+b&=-29.5\\
4a+2b&=1869.1
\end{aligned}\end{displaymath}

We will solve the system the substitution method. Solving the first equation for $ a$ yields $ a=-b-29.5$. Substituting this result into the second equation $ 4a+2b=1869.1$ yields

\begin{displaymath}
\begin{aligned}
4(-b-29.5)+2b&=1869.1 \quad {\text {Since}} ...
...118+2b &=1869.1 \\
-2b&=1987.1 \\
b&=-993.55
\end{aligned}\end{displaymath}

Since $ a=-b-29.5$,

\begin{displaymath}
\begin{aligned}
a &=-(-993.55)-29.5 \\
&=964.05
\end{aligned}\end{displaymath}

The quadratic function that models the revenue of the Kellogg Company is $ R(t)=964.05t^2-993.55t+6984.2$, where $ t$ is the number of years since the end of 1999 and $ R(t)$ is the revenue from sales in millions of dollars.

In each of the preceding examples, the quadratic model fit the data well. This is not always the case, as demonstrated in Example 7.


\begin{example}{\bf Determing When a Quadratic Model Should Not Be Used} \\
The...
...y-to-eat and ready-to-cook breakfast cereal is shown in Table 2.5.
\end{example}


表 格 2.5:
Years Cereal Consumption
Since 1980 (pounds)
$ (t)$ [$ C(t)$]
0 12
1 12
2 11.9
3 12.2
4 12.5
5 12.8
6 13.1
7 13.3
8 14.2
9 14.9


Years Cereal Consumption
Since 1980 (pounds)
$ (t)$ [$ C(t)$]
10 15.4
11 16.1
12 16.6
13 17.3
14 17.4
15 17.1
16 16.6
17 16.3
18 15.6
19 15.5

Source: Statistical Abstract of the United States, 2001, Table 202, p.129.

Explain why you do or not believe that a quadratic function will model the data set well.

解: We first draw the scatter plot of the data set (Figure 2.14).

圖 2.15:
\includegraphics[width=2.4in]{F2-14.eps}

(Note: We have adjusted the viewing window so that we may better analyze the data.)

Recall that a parabola is either concave up everywhere or concave down everywhere. This scatter plot appears to be concave up between $ t=1$ and $ t=10$, concave down between $ t=10$ and $ t=17$, and concave up between $ t=17$ and $ t=19$. The fact that the scatter plot changes concavity causes us to doubt that a quadratic model will fit the data well. In short, the scatter plot doesn't look like a parabola or a portion of a parabola.

The quadratic model that best fits the data is $ C(t)=-0.0179t^2+0.637t+10.8$ and is shown in Figure 2.15.

圖 2.16:
\includegraphics[width=2.4in]{F2-15.eps}

The coefficient of determination ($ r^2=0.853$) is not as close to 1 as in our previous examples. Additionally, we can see visually that the model does not fit the data well.

Once a model has been created, we can often use it to make educated guesses about what may happen in the future. Forecasting the future is a key function for many business.


\begin{example}{\bf Using Quadratic Regression to Forecast Fuel Consumption}\\
...
...umption of these types of vehicles has also
increased (Table2.6).
\end{example}


表 格 2.6: Motor Fuel Consumption of Vans, Pickups, and SUVs
Years Fuel Consumption
Since 1980 (billions of gallons)
$ (t)$ $ F$
0 23.8
5 27.4
10 35.6
15 45.6
19 52.8

Source: Statistical Abstract of the United States, 2001, Table 1-105, p.691.

Model the fuel consumption with a quadratic function and forecast the year in which fuel consumption will reach 81.0 billion gallons.

解: Using quadratic regression, we determine the model to be

$\displaystyle F(t)=0.0407t^2+0.809t+23.3\quad {\text {billion}} \quad {\text {gallons}}$

where $ t$ is the number of years since the end of 1980. We want to know at what value of $ t$ does $ F(t)=81.0$. This problem may be solved algebraically or graphically using technology. We will solve the problem twice (once with each method) and allow you to use the method of your choice in the exercise.

ALGEBRAIC SOLUTION

\begin{displaymath}
\begin{aligned}
81.0 &=0.0407t^2+0.809t+23.3\\
0 &=0.0407t^2+0.809t-57.7
\end{aligned}\end{displaymath}

This is a quadratic function with $ a=0.0407$, $ b=0.809$, and $ c=-57.7$.

Recall that the solution to a quadratic equation of the form $ at^2+bt+c=0$ is given by Quadratic Formula,

$\displaystyle t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting our values of $ a$, $ b$, and $ c$ into the Quadratic Formula yields

\begin{displaymath}
\begin{aligned}
t &=\frac{-0.809\pm \sqrt{(-0.809)^2-4(0.04...
...0}}{0.0814} \\
& =\frac{-0.809\pm 3.17}{0.0814}
\end{aligned}\end{displaymath}

In the context of the problem, we know that $ t$ must be nonegative. Consequently, we will calculate only the nonegative solution.

\begin{displaymath}
\begin{aligned}
t&=\frac{-0.809+3.17}{0.0814}\\
&=\frac{2.36}{0.08414}\\
&=29.0
\end{aligned}\end{displaymath}

We anticipate that at the end of 2009 (29 years after the end of 1980), the fuel consumption will have reached 81.0 billion gallons.

GRAPHICAL SOLUTION
We graph the function $ F(t)=0.0407t^2+0.0809t+23.3$ and the horizontal line $ y=81.0$ simultaneously (Figure 2.16). $ F(t)=81.0$ at the point at which these two functions intersect.

圖 2.17:
\includegraphics[width=2.4in]{F2-16.eps}

Using the intersect feature on the graphing calculator, we determine that the point of intersection is $ (29,81.0)$. The interpretation of the solution is the samr as that given in the discussion of the algebraic method.

2.1 Summary
In this section, you learned how to use quadratic regression to model a data set. You also discovered the importance of analyzing a mathematical model before using it to calculate unknown values.


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