We will start by determining the type of function that best models the data. Then we will be able to write an equation that models the data.
Does the graph of the data suggest a type of function to use?
A graph of the data can suggest a type of function to use. By looking at it we can eliminate possibilities and find the appropriate type of function quicker. Let's graph the given data!
The graph curves and does not look like a graph of an . It may be modeled by a . Note that the numbers of the days have a common difference of 1. Therefore, we will check the data for constant second differences.
| Day
|
Visitors
|
First Differences
|
Second Differences
|
| 1
|
52
|
|
|
| 2
|
197
|
+145 ↩
|
|
| 3
|
447
|
+250 ↩
|
+105 ↩
|
| 4
|
805
|
+358 ↩
|
+108 ↩
|
| 5
|
1270
|
+465 ↩
|
+107 ↩
|
The second differences oscillate between 105 and 108, so we can assume they are close to being constant. Therefore, a quadratic model fits the data.
Will your equation fit the data exactly? How do you know?
Since the second differences are only almost constant, our equation will not fit the data exactly. We will still find the equation of a quadratic function that models the data!
y= ax^2+ bx+ c
To write our equation, we have to determine the values of a, b, and c. We will substitute three of the given points into the of a quadratic equation. Doing so will create a that we can solve for the values of a, b, and c. Let's start with (1,52).
y=ax^2+bx+c
52=a( 1)^2+b( 1)+c
a+b+c=52
We just wrote our first equation! Now let's do the same thing for (2,197).
y=ax^2+bx+c
197=a( 2)^2+b( 2)+c
197=a(4)+b(2)+c
197=4a+2b+c
4a+2b+c=197
To find our third and last equation we will use (3,447).
y=ax^2+bx+c
447=a( 3)^2+b( 3)+c
447=a(9)+b(3)+c
447=9a+3b+c
9a+3b+c=447
We now have a system of three equations.
a+b+c=52 & (I) 4a+2b+c=197 & (II) 9a+3b+c=447 & (III)
Let's solve this system using the . We will start by subtracting Equation (I) from Equation (II) to eliminate the c-variable.
a+b+c=52 4a+2b+c=197 9a+3b+c=447
a+b+c=52 4a+2b+c-( a+b+c)=197- 52 9a+3b+c=447
a+b+c=52 4a+2b+c-a-b-c=197-52 9a+3b+c=447
a+b+c=52 3a+b=145 9a+3b+c=447
Now let's subtract Equation (I) from Equation (III) to eliminate the c-variable once more.
a+b+c=52 3a+b=145 9a+3b+c=447
a+b+c=52 3a+b=145 9a+3b+c-( a+b+c)=447- 52
a+b+c=52 3a+b=145 9a+3b+c-a-b-c=447-52
a+b+c=52 3a+b=145 8a+2b=395
Now neither Equation (II) nor Equation (III) includes the c-variable. These equations form a system with only two variables, a and b. Let's solve this system by using the Elimination Method again. Since neither variable has the same or opposite , we will need to multiply one equation by a number first.
a+b+c=52 3a+b=145 8a+2b=395
â–¼
(II), (III):Solve by elimination
a+b+c=52 6a+2b=290 8a+2b=395
a+b+c=52 6a+2b=290 8a+2b-( 6a+2b)=395- 290
a+b+c=52 6a+2b=290 8a+2b-6a-2b=395-290
a+b+c=52 6a+2b=290 2a=105
a+b+c=52 6a+2b=290 a=52.5
a+b+c=52 6( 52.5)+2b=290 a=52.5
a+b+c=52 315+2b=290 a=52.5
a+b+c=52 2b=- 25 a=52.5
a+b+c=52 b=- 12.5 a=52.5
Finally, to find the value of c we will substitute a=52.5 and b=- 12.5 into Equation (I).
a+b+c=52 b=- 12.5 a=52.5
52.5+( - 12.5)+c=52 b=- 12.5 a=52.5
52.5-12.5+c=52 b=- 12.5 a=52.5
40+c=52 b=- 12.5 a=52.5
c= 12 b=- 12.5 a=52.5
Since we are only modeling the data we can round the values of a and b to the nearest .
a≈ 53 and b≈ - 13
Finally let's substitute the values a, b, and c into the standard form of a quadratic equation.
y= 53x^2+( - 13)x+ 12 ⇔ y=53x^2-13x+12
Below we have included a graph that shows how the equation models the given data.
