$29$ . Multiple choice - Bushels of corn exported from the U.S., in millions, from $2008$ to $2012$ , can be modeled by a quadratic function. In $2008$ , the U.S. exported approximately $1849$ million bushels of com. In $2009$ , the U.S. exported approximately $1979$ million bushels of corn, which was a maximum for that time period. According to the above information, which of the following best approximates the U.S. corn exports in $2012$ ? $\newline$ A. $809$ million bushels $\newline$ B. $1170$ million bushels $\newline$ C. $1589$ million bushels $\newline$ D. $2369$ million bushels

View step-by-step help

Home

Math Problems

Algebra 1

Interpret regression lines

Full solution

29

. Multiple choice - Bushels of corn exported from the U.S., in millions, from

2008

2012

, can be modeled by a quadratic function. In

2008

, the U.S. exported approximately

1849

million bushels of com. In

2009

, the U.S. exported approximately

1979

million bushels of corn, which was a maximum for that time period. According to the above information, which of the following best approximates the U.S. corn exports in

2012

\newline

809

million bushels

\newline

1170

million bushels

\newline

1589

million bushels

\newline

2369

million bushels

Understand Given Information: Understand the given information and what is being asked. We know that the exports can be modeled by a quadratic function, and we have data for two years: $2008$ and $2009$ . The exports in $2008$ were $1849$ million bushels, and in $2009$ , which was a maximum, the exports were $1979$ million bushels. We need to find the best approximation for the exports in $2012$ .
Find Vertex of Quadratic Function: Since $2009$ was a maximum, the vertex of the quadratic function will be at the point $(2009, 1979)$ . We can use the vertex form of a quadratic function, which is $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola. In this case, $h = 2009$ and $k = 1979$ .
Calculate Value of 'a': Use the information from $2008$ to find the value of 'a'. We know that in $2008$ (which is one year before the maximum), the exports were $1849$ million bushels. Plugging this into the vertex form, we get $1849 = a(2008 - 2009)^2 + 1979$ . Simplifying, we get $1849 = a(-1)^2 + 1979$ , which leads to $a = 1849 - 1979$ .
Determine Quadratic Function: Calculate the value of $a$ . $a = 1849 - 1979 = -130$ . Now we have the value of $a$ , and our quadratic function is $y = -130(x - 2009)^2 + 1979$ .
Find Exports in $2012$ : Use the quadratic function to find the exports in $2012$ . We substitute $x = 2012$ into the function: $y = -130(2012 - 2009)^2 + 1979$ . This simplifies to $y = -130(3)^2 + 1979$ .
Find Exports in $2012$ : Use the quadratic function to find the exports in $2012$ . We substitute $x = 2012$ into the function: $y = -130(2012 - 2009)^2 + 1979$ . This simplifies to $y = -130(3)^2 + 1979$ .Calculate the value of $y$ for $x = 2012$ . $y = -130(9) + 1979 = -1170 + 1979 = 809$ . According to our calculation, the exports in $2012$ would be $809$ million bushels.