To solve this exercise, we need to understand the given model. Then, we will shift the model to the right and sketch both graphs on the same coordinate plane.

Understanding the Model

We are told that our distance

d

in miles from the halfway point can be modeled by the function

d = 72 ∣ x - 30 ∣,

where

x

is the time in days. Therefore, when

d = 0,

we have reached the halfpoint. Let's calculate

x

for

d = 0 .

d = 72 ∣ x - 30 ∣

Substitute

$d = 0$

0 = 72 ∣ x - 30 ∣

▼

Simplify right-hand side

DivEqn

$LHS / 72 = RHS / 72$

0 = ∣ x - 30 ∣

$∣ a ∣ = 0 \Leftrightarrow a = 0$

0 = x - 30

AddEqn

$LHS + 30 = RHS + 30$

30 = x

RearrangeEqn

Rearrange equation

x = 30

We found that we will by halfway through after

30

days. Thus, we will travel

4320

miles in

2 \cdot 30 = 60

days. Since

x = 0

represents June

1,

the trip will finish on July

30 .

Shifting the Model

Suppose now that our plans are altered so that the model is shifted

10

units to the right. We have a new formula to model the distance

d

from the halfway point.

d = 72 ∣ (x - 10) - 30 ∣ \Leftrightarrow d = 72 ∣ x - 40 ∣

Let's graph the original and the new model on the same coordinate plane. Since the trip is

4320

miles, the maximum distance from the halfway point is

4320 \div 2 = 2160

miles. Therefore, the maximum value for both functions is

2160 .

The new model delays the trip by $10$ days. It does not modify the length or the duration of the trip. With this model, $x = 0$ represents June $11 .$ In general, a right shift of $h$ units delays the trip in $h$ days, and $x = 0$ represents June $1 + h .$

Exercise