Question

The table shows the relationship between the diameter in kilometers of an oil spill and the time in days. A quadratic function can be used to model this relationship.What is the best prediction of the time required for the oil spill to reach a diameter of 10 km?

Answer 1

We can use the general form of a quadratic model:

$y=ax^2+bx+c$

We can find the parameters a, b and c by taking some pair of values from the table, like this:

Let's take the pair (0,0), if we replace these values for y and x into the above model we get:

$\begin{gathered} 0=a\times(0)^2+b\times(0)+c \\ 0=c \\ c=0 \end{gathered}$

Now let's take the pair (1, 2.4), by replacing these values into the above model for x and y, we get:

$\begin{gathered} 2.4=a\times(1)^2+b\times(1) \\ 2.4=a+b \end{gathered}$

From this equation, we can solve for a to get:

$\begin{gathered} 2.4=a+b \\ 2.4-b=a+b-b \\ a=2.4-b \end{gathered}$

By taking the pair (2, 9.4) we get:

$\begin{gathered} 9.4=a(2)^2+b\times2 \\ 9.4=4a+2b \end{gathered}$

We can replace the expression that we got previously a = 2.4 - b into the above equation to get:

$9.4=4(2.4-b)+2b$

From this expression, we can solve for b like this:

$\begin{gathered} 9.4=4(2.4-b)+2b \\ 9.4=9.6-4b+2b \\ 9.4-9.6=9.6-9.6-2b \\ -0.2=-2b \\ -2b=-0.2 \\ b=\frac{-0.2}{-2} \\ b=0.1 \end{gathered}$

Now we can replace it into the equation a = 2.4 - b, then we get:

a = 2.4 - 0.1 = 2.3

Then we get the expression:

$y=2.3x^2+0.1x$

Now we just have to evaluate 10 km into the equation, then we get:

$y=2.3\times(10)^2+0.1\times(10)=231$

Then, the best prediction of the time required for the oil spill to reach a diameter of 10 km is 233 days