Hello!

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

slope-m, y-intercept(b)

m=(y2-y1)/(x2-x1)=(3000-14000)/(5-1)= (-11000)/(4)=-2750

point(1,14000) -> y=-2750(x-1)+14000

y=-2750x+16750

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)

equation: y=a(x-1)^2+14000

(5,3000) for x,y

3000=a(5-1)^2+14000

3000=a*16+14000

-11000=16a

a=-687.5

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?

Thank you.

Laura