Exponential Growth Rates

Question

So if you are given two different numbers to determine a growth rate, do you use to largest number compared to the value when x=0. For example the problem I am working on is: Your grandfather purchased a house for $45,000 in 1952 and it has increased in value according to a function y = v(x), where x is the number of years owned. These questions probe the future value of the house under various mathematical models.

Suppose the value of the house is $60,000 in 1962 and $140,000 in 1967. Assume v(x) is a quadratic function. Find a formula for v(x). If I take (60000/45000)^1/10 I get a growth rate of 1.03 If I take (140000/45000)^1/15 I get a growth rate of 1.08 And if I take (140000/60000)^1/5 I get a growth rate of 1.18 Which of these am I supposed to use?

Answer 1

Assuming the function you are looking for is quadratic (as you suppose), then $v(x)=ax^2+bx+c$ satisfies $v(0)=45000$ (we take as year $0$ the year $1952$), $v(10)=60000$ and $v(15)=140000.$

You have to solve the system

$$\left\{ \begin{array}{ccccccc} a \cdot 0^2 & + & b \cdot 0 & + & c & = & 45000\\ a \cdot 10^2 & + & b \cdot 10 & + & c & = & 60000 \\ a \cdot 15^2 & + & b \cdot 15 & + & c & = & 140000 \end{array} \right.$$

which solution is $a=2900/3, b=-24500/3, c=45000.$ That is,

$$v(x)=\displaystyle \frac{2900}{3}x^2-\frac{24500}{3}x+45000.$$

You are not supposed to use any specifically ratio change, if you assume quadratically behaviour, because the ratio change varies. You have to evaluate at the convenient $x$ (always assuming quadratic behaviour).

Answer 2

Once you decided about the model, it is quite simple because you have very few data points. In the case of a quadratic function, building and solving the three equations, you obtained what Manuel Fdz Lpz answered. Where the problem would come is when you will use the model for extrapolation. For example, using the quadratic equation, you would predict a value of $268,333$ at year $20$ and $670,000$ at year $30$.

You could consider also, and this is another possible model, that it is the value of an investment and find the interest rate. In this case, the model would be $$v(x)=45000 (1+r)^x$$. Notice that we use the initial value and only one extra parameter (while we have two extra data points). Adjusting the model requires regression (since we have more equations or data points than parameters). Applied to your data, the model would then be $$v(x)=45000 \times 1.0711^x$$ Using this exponential equation, you would predict a value of $177,744$ at year $20$ and $353,253$ at year $30$ which is probably more realistic that what gives the quadratic model. For sure, this last model predicts a value of $89,434$ for $x=10$ (this is higher than the $60,000$ given and a value of $126,081$ for $x=15$ (this is lower than the $140,000$ given).

If you want to dicriminate between models, what would be good is to have more data points. If you are able to get other data points, please post them and I shall look at them for you.