A$16,000 robot depreciates linearly to zero in 10 years. (a) find a formula for its value as a function of time, t, in years.

The equation become y = -1600x + 16000.

It is a system of an equation in which the highest power of the variable is always 1. A one-dimension figure that has no width. It is a combination of infinite points side by side.

Given

A $16,000 robot depreciates linearly to zero in 10 years.

A formula for its value as a function of time, t, in years.

Let y be the cost of the robot and x be the time.

We know the equation of the line is y = mx + c

At the time of launch, its cost is $16,000 so the point becomes (0,16000) and after 10 years its cost becomes zero then the point becomes (10,0).

For the first point (0, 16000)

16000 = 0(x) + c

        c = 16000

For second point (10,0)

0 = 10(m) + 16000

m = -1600

Thus the equation become y = -1600x + 16000.

More about the linear system link is given below.

link

We are given, cost of the robot for 0 number of year = $16,000.

0 represents initial time of the robot.

After 10 year cost of the robot is = $0

The problem is about the number of the years and cost of the robot over different number of years.

So, we could take x coordinate by number of hours and y coordinate for y number of hours.

So, from the problem, we could make two coordinates for the given situation.

(x1,y1) = (0, 16000) and (x2,y2) = (10, 0).

In order to find the function of time, we need to find the rate at which robot rate depreciates each year.

Slope is the rate of change.

So, we need to find the slope of the two coordinates we wrote above.

We know, slope formula

Plugging values in formula, we get

Brecasue of depreciation we got a negative number for slope or rate of change.

Therefore, rate of depreciation is $1600 per year.

We already given inital cost, that is $16,000.

So, we can setup an a function

f(x) = -1600x + 16000.

But the problem is asked to take the variable t for time.

Replacing x by t, we get

f(t) = -1600t + 16000.