Chapter 7.6, Problem 62PPS

a.

To determine

To describe: The graph of $g\left(x\right)$ when she invests $1000 and when she invests $7500 in her bank account.

Explanation of Solution

Given:

The person L’s saving can be modeled by the function $g\left(x\right)=P{\left(1.002\right)}^t$ where $P$ is the initial amount and $t$ is time in years.

The two functions for her investment can be written as $g\left(x\right)=1000{\left(1.002\right)}^t$ and $g\left(x\right)=7500{\left(1.002\right)}^t$ .

Observe that $g\left(x\right)=7.5\left[1000{\left(1.002\right)}^t\right]$ . That means, the graph of the investment $1000 get stretched in y-direction by a factor $7.5$ to get the graph of the investment of $7500.

Now, graph the two function to verify the same.

  

b.

To determine

To find: the affect of the saving of $250 on the graph of $g\left(x\right)$ .

Answer to Problem 62PPS

The graph of $g\left(x\right)$ gets shifted 250 units upward

Explanation of Solution

Given:

The person L’s keeps $250 cash in her apartment which she considers as her saving.

Since the amount $250 is not deposited in the bank, it will not earn any interest. So, it will not grow by time.

Therefore, her total saving can be modeled by the function $g\left(x\right)=P{\left(1.002\right)}^t+250$ .

That means, the graph of $g\left(x\right)$ gets shifted 250 units upward.