The price of a home in Medford was dollar-sign Baseline 100,000 in 1985 and rose to dollar-sign 148 ,000 in 1997.
a. Create two models, f(t) assuming linear growth and g(t) assuming exponential growth, where t is the number of years after 1985.
Round coefficients to three decimal places when necessary.
For the model for f(t) I got 100000+4000t and for g(t) I got 100000*(1.48)^1/t.
After this I would plug in 12 to t and get my answer.
But it says I'm wrong please help.
The linear model is ok, but the exponential growth is way off.
It grew by a factor of 1.48 in 12 years, so
g(t) = 100000 * 1.48^(t/12)
Clearly you did not try your t=12 input in g(x)
exponential functions are "never" b^(k/t)
as t grows large, that approaches just 1.
To create the linear growth model, f(t), you need to find the slope of the line passing through the two points (1985, $100,000) and (1997, $148,000).
Using the formula for the slope, m = (y2 - y1) / (x2 - x1), where (x1, y1) = (1985, $100,000) and (x2, y2) = (1997, $148,000), we can calculate:
m = (148,000 - 100,000) / (1997 - 1985) = 48,000 / 12 = 4,000.
Since the y-intercept is given as $100,000 in 1985, the equation for the linear growth model is:
f(t) = 100,000 + 4,000t.
For the exponential growth model, g(t), we need to find the value of the growth rate factor, r. This can be done by calculating the ratio of the final value ($148,000) to the initial value ($100,000):
r = ($148,000) / ($100,000) = 1.48.
The general formula for exponential growth is g(t) = a * (1 + r)^t, where a is the initial value. Plugging in the values, we get:
g(t) = $100,000 * (1.48)^t.
Now, to evaluate both models at t = 12, you can substitute the value of t into the equations:
f(12) = 100,000 + 4,000 * 12 = $148,000.
g(12) = 100,000 * (1.48)^12 = approximately $281,672.
Therefore, the price of a home in Medford, according to the linear growth model, is $148,000 after 12 years, while the exponential growth model predicts a price of approximately $281,672 after 12 years.