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

Answer:

y = -3250t + 25000

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 =$\frac{y^{2}-y 1}{t 2-t 1}$

m =$\frac{12000-25000}{4-0}$

m = -  $\frac{13000}{4}$= -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

Answer:

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

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

$m=\frac{y2-y1}{t2-t1}$

substitute the values

$m=\frac{12,000-25,000}{4-0}$

$m=\frac{-13,000}{4}$

$m=-\ 3,250\ per\ year$ ---> 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

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