Question

Tonya and Leo each bought a cell phone at the same time. The trade-in values, in dollars, of the cell phones are modeled by the given functions, where x is the number of months that each person has owned the phone. Tonya's Phone Leo's Phone f(x) =490(0.88)x x g(x) 0 480 2 360 4 270 phone has the greater initial trade-in value. During the first four months, the trade-in value of Tonya's phone decreases at an average rate the trade-in value of Leo's phone.

Answer 1

The answer is Tonya's phone had the greater initial trade-in value.

Leo's phone decreases at an average rate slower than the trade in value of Tonya's phone.

Explanation:

Given

Tonya

`f(x) = 490 * 0.88^x`

Leo

`x \to g(x)`

`0 \to 480`

`2 \to 360`

`4 \to 470`

Solving (a): The phone with greater initial value

The initial value is when x = 0. So, we have:

`f(x) = 490 * 0.88^x`

`f(0) = 490 * 0.88^0`

`f(0) = 490 * 1`

`f(0) = 490`

From Leo's table

`g(0) = 480`

By comparison;

`f(0) > g(0)`

i.e.

`490 > 480`

So: Tonya's had the greater initial trade-in value

Solving (b): The phone with lesser rate

An exponential function is:

`y = ab^x`

Where:

`b \to rate`

For Tonya

`b = 0.88`

For Leo, we have:

`(x_1,y_1) = (0,480)`

`(x_2,y_2) = (2,360)`

So, the equation becomes:

`y = ab^x`

`480 = ab^0` and
`360 = ab^2`

Solving
`480 = ab^0`, we have:

`480 = a * 1`

`480 = a`

`a= 480`

`360 = ab^2` becomes

`360 = 480 * b^2`

Divide both sides by 480

`0.75 = b^2`

Take square roots

`0.87 = b`

`b=0.87` -- Leo's rate

By comparison; Leo's rate is slower i.e. 0.87 < 0.88