car 1 is exponential
car 2 is linear
note that dividing any value of car 1 over its previous year value leads to the same result. for example, (year2 value)/(year1 value) = 16346.60/17390 = 0.94; and also (year3 value)/(year2 value) = 15365.80/16346.60 = 0.9399997553008 which rounds to 0.94. the fact we get the same result each time suggests that we have an exponential model.
the value of car 2 drops by 1000 each time. this constant drop is why it's a linear function. put more technically, the slope is rise/run = -1000/1 = -1000. the negative rise means it's actually a fall in value.
equation for car 1 is f(x) = 18500(0.94)^x
equation for car 2 is f(x) = -1000x + 18500
in part a, we found that the multiplier was 0.94 by dividing any given value by its previous one. so this is the base of the exponential function, or the value of b. the value of 'a' is a = 18500 which is the starting amount. so we go from y = a*b^x to y = 18500*(0.94)^x. this is the equation for car 1.
now onto car 2's equation. in part a, we found car 2's value decrease at a constant rate of 1000 per yer. this was mentioned to be the slope, so m = -1000. the y intercept is the starting value so b = 18500. we go from y = mx+b to y = -1000x+18500
car 1's value after 9 years is $10,600.40
car 2's value after 9 years is $9,500.00
there isn't much of a significant difference in car value. however, i'm not sure what your teacher considers to be "significant". despite that, baxter is better off going with car 1 because it has higher value compared to car 2 after 9 years have passed.
plug x = 9 into each function. i'm going to make f(x) be the function for car 1 while g(x) be the function for car 2, to make comparison clearer.
f(x) = 18500(0.94)^x
f(9) = 18500(0.94)^9
f(9) = 18500(0.57299480222861)
f(9) = 10,600.4038412292
f(9) = 10,600.40
g(x) = -1000x+18500
g(9) = -1000(9)+18500
g(9) = -9000+18500
g(9) = 9,500.00