Using the quadratic function to make a prediction

Question

I've been given a simple problem: use the quadratic function to predict the number of subscribers expected in 2010.

According to the table, $$\begin{array}{|c|c|} \hline \text{Year} & \text{Subscribers (millions)} \\ \hline 1985 & 1 \\ \hline 1990 & 11 \\ \hline 2000 & 741\\ \hline \end{array}$$

I was told to set-up the series of equations like $$A(1985^2) + B(1985) + C = 1$$and so on. However, when I do that and put them in a matrix to solve, I get an answer that is far from correct. Any help in getting me on the right track? I don't know why I'm struggling with this so much... That said, I do suspect I was not told how to set-up this problem correctly.

Answer 1

did you try the system ...

$$A(1985^2) + B(1985) + C = 1 \tag 1 $$

$$A(1990^2) + B(1990) + C = 11 \tag 2 $$

$$A(2000^2) + B(2000) + C = 741 \tag 3 $$

$$ (2)-(1) \implies A(1990^2-1985^2) +5B = 10 \tag 4$$

$$ (3)-(1) \implies A(2000^2-1985^2) +15B = 740 \tag 5$$

now the equation generated from $(5)-3\times (4)$ can be solved for A , sub into (4) and (1) to get B and C

Answer 2

To simplify things we take $1985$ as the origin of time. Let's determine the quadratic function $f(t) = at^2 + bt + c$ such that $f(0) = 1$, $f(5) = 11$ and $f(15) = 741$. Its coefficients $a$, $b$ and $c$ verify $$ \left\{ \begin{alignedat}{4} c &= 1 & \quad (1)\\ 25a {}+{} 5b &= 10 & \quad (2) \\ 225a {}+{} 15b &= 740. & \quad (3) \end{alignedat} \right. $$ Doing $(3) - 3\times (2)$ gives $a = 71/15$, and $(3) - 9\times (2)$ gives $b = -65/3$. Thus $$ f(t) = \frac{71}{15} t^2 - \frac{65}{3} t + 1. $$

Answer 3

If we name $S$ the number of subscribers and $Y$ the year, you are asked to find $A,B,C$ in the model $$S=A\,Y^2+B\,Y+C$$ and use the values in the table to find the coefficients.

To avoid problems with large numbers, you could rewrite the model as $$S=A\,t^2+B\,t+C$$ where $t=Y-1985$. In such a case, using the same idea as WW1, the equations write $$A (0^2)+B(0)+C=1$$ $$A (5^2)+B(5)+C=11$$ $$A (15^2)+B(15)+C=741$$ which is much simpler to solve (in particular thanks to the first equation).

Solve for $A,B,C$ and compute the value of the expression for $t=2010-1985=25$.

I am sure that you can take from here.