I need to solve this problem and I am not sure if I did it right.
"The value of an automobile over time is given in the following table:
1 | 14000
2 | 9100
3 | 6200
4 | 4000
5 | 3000
a) Use graphing software or an algebraic method to determine an equation that fits
i) the linear model
ii) the quadratic model
iii) the exponential model
b) Use each model to predict the value of the car after 10 years
c) Which result is most reasonable? Give reasons for your answer
d) Which function provides the best model? "
a) Linear: y=mx+b
point(1,14000) -> y=-2750(x-1)+14000
Quadratic: y=a(x-p)^2 + q
p and q-coordinates of the vertex
(1,14000) is the highest point on the graph -> (p,q)
(5,3000) for x,y
The quadratic equation: y=-687.5(x-1)^2+14000
Exponential: y=ar^x, where r is the constant ratio
Ratio: 9100/14000=0.65 ; 6200/9100=0.68 ; 4000/6200=0.65 ; 3000/4000=0.75
Average: 0.6825 ---> r=0.68
y=a(0.68)^x with y=14000 when x=1 ---> 14000=a(0.68)^1 or a=14000/0.68=20.59
The exponential equation is: y=21(0.68)^x
b) After 10 years ...Linear:y= -10750 , Quadratic: y= -41687.5 , Exponential: y=0.44
c)Exponential is the most reasonable based upon the 10th year values.
Did I solved it correct? Are there any mistakes?
Maybe a 10 year old toy car ?
Thanks. Calculation error.
The exponential equation is: y=20588.24(0.68)^x
b) After 10 years ...
[...] Exponential: y=435.22
Now it's ok?!
Can you give me a little hit for d ? I think (maybe I'm wrong) that the best model is the exponential one? Points c and d seems similar to me.
The general equation of a quadratic function is:
Use 3 of the "points" (value of car over time) and plug in the the year(for x) and the value(for y) to get a system of 3 simultaneous equations in (a,b,c). For instance:
Solve for (a,b,c). You should come out with:
To do d) I would graph the "points" and the graphs of the 3 functions and would try to judge which function describes best the development of the car value over time.