Question

bought a new car in 2011 for 25000. in 2015, joe was offered a fair price of 12000 for his car but he turned it down. build a linear algebraic model, that helps joe find the car's value when it is t years old

Answer 1

y = -3250t + 25000

Step-by-step explanation:

Assuming the years between 2011-2015 to be a straight line

We know that, for a straight line, y = mx + c,

Where, m is slope of unit rate

c is the y-intercept

Consider t to be the no. of years since 2011 and y to be the car’s value

So, for 2011, t=0

And for 2015, t=4 (2015-2011)

Finding the slope m

m =
`(y^(2)-y 1)/(t 2-t 1)`

m =
`(12000-25000)/(4-0)`

m = -
`(13000)/(4)`= -3250 per year (negative because it is a decreasing function)

We have, c = 25000

Substituting the values in the straight line equation,

y = -3250t + 25000

Answer 2

`y=-3,250t+25,000`

Step-by-step explanation:

we know that

The linear equation in slope intercept form is equal to

`y=mx+b`

where

m is the slope of unit rate of the linear equation

b is the y-intercept or initial value of the linear equation

Let

t ----> the number of years since 2011

y --->the car's value

In this problem the year 2011 represent t=0

so

the year 2015, represent t=4 years (2015-2011)

we have the ordered pairs

(0,25,000) ----> represent the y-intercept

(4,12,000)

Find the slope m

The formula to calculate the slope between two points is equal to

`m=(y2-y1)/(t2-t1)`

substitute the values

`m=(12,000-25,000)/(4-0)`

`m=(-13,000)/(4)`

`m=-\$3,250\ per\ year` ---> is negative because is a decreasing function

we have

`b=\$25,000` ----> value of y when the value of x is equal to zero (initial value)

substitute the given values

`y=-3,250t+25,000`