Given a table of values, how do you know a quadratic model is the most appropriate one?
See solution.
Practice makes perfect
Let's suppose we are given a table showing the depth, in millimeters, of the snow in a small town in Ohio from 6A.M. to 11A.M.
Let's calculate the first and second differences to see if the situation can be modeled by a quadratic function.
The first differences are not constant. Therefore, a linear model is not the most appropriate one for the situation. However, the second differences are constant and thus a quadratic model can be used to describe the situation. Let d(t) be the depth of the snow, in millimeters, t hours after 6 A.M.
d(t)=at^2+bt+c
Since at 6A.M. we have that t=0 and d(t)=6, we can substitute 0 and 6 for t and d(t), respectively, in the above equation to find the value of c.
Now that we know that c= 6, we can partially write the quadratic function.
d(t)=at^2+bt+ 6
By looking at the table, we can see that at 7A.M. — 1 hour after 6A.M. — the depth of the snow is 17 millimeters. Similarly, at 8A.M. — 2 hours after 6A.M. — the depth of the snow is 34 millimeters.
Values for t and d(t)
d(t)=at^2+bt+6
Simplification
t= 1, d(t)= 17
17=a( 1)^2+b( 1)+6
a+b=11
t= 1, d(t)= 34
34=a( 2)^2+b( 2)+6
4a+2b=28
Note that we obtained two equations in a and b. We can write and solve a system of equations.
a+b=11 & (I) 4a+2b=28 & (II)
To solve the system we will use the Substitution Method. Let's start by isolating a in Equation (I). This is done by subtracting b on both sides.
a+b=11 4a+2b=28 ⇔ a=11-b 4a+2b=28
Now, we can substitute 11-b for a in Equation (II).
